7350399 Problems and Solutions in Mathematics Li Ta Sien World Scientific

PROBLEMS AND

SOLUTIONS I N

MATHEMATICS

Major American Universities Ph.D. Qualifying Questions and Solutions

PROBLEMS AND

SOLUTIONS IN

MTHEMTICS Compiled by:

Chen Ji-Xiu, Jiang Guo-Ying, Pan Yang-Lian, Qin Xe-Hu, Tong Yu-Sun, Wu Quan-Shui and Xu Sheng-Zhi

Edited by:

LI TA-TSIEN Fudm University

orld Scientific NewJersey London Singapore Hong Kong

Published by

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Library of Congress Cataloging-in-Publication Data Problems and solutions in mathematics I [edited by] Li Ta Tsien.

p. cm. Includes bibliographical references. ISBN 9810234791 -- ISBN 9810234805 (pbk) 1. Mathematics -- Problems, exercises, etc. I. Li, Ta-chen.

QA43.P754 1998 510,76--dc21 98-22020

CIP

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First published 1998 Reprinted 2002,2003

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V

PREFACE

This book covers six aspects of graduate school mathematics: Algebra, Topology, Differential Geometry, Real Analysis, Complex Analysis and Par- tial Differential Equations. It contains a selection of more than 500 problems and solutions based on the Ph.D. qualifying test papers of a decade of in- fluential universities in North America. The mathematical problems under discussion are kept within the scope of the textbooks for graduate students. Finding solutions to these problems, however, involves a deep understanding of mathematical principles as well as an acquisition of skills in analysis and computation. As a supplement to textbooks, this book may prove to be of some help to the students in taking relevant courses. It may also serve as a reference book for the teachers concerned.

It has to be pointed out that this book should not be regarded as an all- purpose troubleshooter. Nor is it advisable to take the book as an exemplary text and commit to memory all the problems and solutions and make an indis- criminate use of them. Instead, the students are expected to make a selective survey of the problems, take a do-it-yourself approach and arrive at their own solutions which they may check against those listed in the book. It would be gratifying to see that the students can work out the problems on their own and come up with better solutions than those provided by the book. If the students fail to do so or their solutions may turn out to be incomplete, it may reveal the inadequacy of their knowledge or approach, thus spurring them to greater efforts to promote their skills. The very purpose of the authors in writing the book is just to help the students to discover the truth by trial and error.

This book was inspired by Professor K. K. Phua’s proposals. We are particularly grateful to him for his support. We also wish to thank Dr. Xu Pei- jun, Professors Zhang Yin-nan, Hong Jia-Xing and Chen Xiao-man for their painstaking efforts to collect test-oriented data. For selecting problems and providing solutions, we wish to acknowledge the following professors respectively: Wu Quan-shui (Part I), Pan Yang-lian (Part II), Jiang Guo-ying (Part

vi

111), Tong Yu-sun, Xu Sheng-zhi (Part IV), Chen Ji-xiu (Part V) and &in Tie-hu (Part VI). We are also indebted to Professor Guo Yu-tao for carefully reading and correcting the manuscript. Finally, we pay tribute to Dr. Cai Zhi-jie for printing out the manuscript.

Li Ta-tsien Department of Mathematics Fudan University Shanghai 200433 China

vii

CONTENTS

................................................................ Preface (v) ......................................................... Part I. Algebra (1)

1. Linear Algebra ................................................... (3) 2. Group Theory ................................................... (26) 3. Ring Theory (44) 4. Field and Galois Theory (59)

Part 11. Topology (81) 1. Point Set Topology .............................................. (83) 2. Homotopy Theory ............................................... (99) 3. Homology Theory .............................................. (118)

Part 111. Differential Geometry ........................................ (151) 1. Differential Geometry of Curves ................................ (153) 2. Differential Geometry of Surfaces (171) 3. Differential Geometry of Manifold (194)

Part IV. Real Analysis ................................................ (229) 1. Measurablity and Measure (231) 2. Integral ........................................................ (256) 3. Space of Integrable Functions (283) 4. Differential ..................................................... (302) 5. Miscellaneous Problems ......................................... (322)

Part V. Complex Analysis ............................................ (333) 1. Analytic and Harmonic Functions (335) 2. Geometry of Analytic Functions ................................ (360) 3. Complex Integration ............................................ (377) 4. The Maximum Modulus and Argument Principles ............... (413) 5. Series and Normal Families ..................................... (433)

Part VI. Partial Differential Equations ................................ (455) 1. General Theory ................................................ (457) 2. Elliptic Equations .............................................. (472) 3. Parabolic Equations ............................................ (496) 4. Hyperbolic Equations ........................................... (513)

Abbreviations of Universities in This Book ............................ (539)

..................................................... .........................................

......................................................

............................... ..............................

......................................

...................................

...............................

Part I

Algebra

3

Section 1 Linear Algebra

1101

Let V be a real vector space of dimension a t least 3 and let T E Endn(V). Prove that there is a non-zero subspace W of V , W # V , such that T ( W ) C W .

Solution, (Indian u)

Make V into an B[X]-module by defining w = T ( v ) for all w E V . Thus

for a,Xi E B[X] and v E V i=O

It is clear that a subspace W of V is an B[X]-submodule of V if and only if

Now suppose V is a simple IR[X]-module. Then V E E[X] / I for some maximal ideal of B[X]. Since R [ X ] is a P.I.D., there exists an irreducible polynomial f (X) of B[X] such that I = (f(X)). So

T ( W ) c W .

3 5 dimR(V) = dimnR[X]/(f(X)) = degf(X).

This implies that we have an irreducible polynomial f ( A ) with degree 2 3 in B[X]. This is a contradiction. Hence V is not a simple B(X)-module, that is, there is a non-zero subspace W of V , W # V , such that T ( W ) W .

1102

Let V be a finite dimensional vector space over a field K . Let S be a linear transformation of V into itself. Let W be an invariant

subspace of V (that is, SW C W ) . Let m(t), ml(t), and rnz(t) be the minimal polynomial of S as linear transformation of V, W and V/W respectively.

(a) Prove that m(t) divides ml(t) . r n 2 ( t ) .

4

(b) Prove that if ml(t) and mZ(t) are relatively prime, then

m(t) = m1(t) * m2(t).

(c) Give an example of a case in which m(t) # ml(t) . mz(t). (Indiana)

Solution. As usual, T/ can be viewed as a K[t]-module via the linear transformation

S . Since W is an S-invariant subspace of V, W is a K[t]-submodule of V. Then it is clear that (m(t)) = A n n ~ p l V ’ , (ml(t)) = A n n ~ p l W and (m2(t)) = A%[t] V l W .

(a) Since

m1(t) . m2(t) . v c m1(t) . W = 0,

~ ( t ) m ( t ) E A n n ~ p l V = (m(t)).

Hence m(t) divides ml(t) m2(t).

(b) Since m(t) E AnnK[t]V G AnnK[t]W = (m1(t>),

rnl(t) divides m(t). Similarly, m2(t) divides m(t). Since ml(t) and mz(t) are relatively prime, ml(t).m2(t) divides m(t). Then we have m(t) = ml(t).m2(t), since m(t), ml(t) and mz(t) are all monic polynomials.

(c) Let W be a 2-dimensional vector space over the field Q of rational numbers and S : W -+ W be a linear transformation with minimal polynomial t2 + 1. Let V = W @ W and S : V --t V be the natural extenaion of S to V. Then it is clear that m(t) = ml(t) = m2(t) = t2 + 1. So m(t) # ml(t) * m2(t)

in this example.

1103

Let V be a finite dimensional vector space over lR and T : V -+ V be a linear transformation such that (a) the minimal polynomial of T is irreducible and (b) there exists a vector u E V such that {Tiv I i 2 0) spans V. Show that V doesn’t have proper T-invariant subspace.

Solution. V can be viewed as a module over the polynomial ring l R [ X ] via f ( X ) . z =

f ( T ) 3 (x), for any f(A) E lR[X] and z E V. Then we have V = lR[X] . v , a cyclic module, since {Taw I i 2 0) spans V by (b). Let m(X) be the minimal

(In d i m a)

5

polynomial of the linear transformation T : V -+ V . Then m(A) E Annwfxl(v). Since m(A) is irreducible, we have

R[X]/(m(A)) E R[A] * 21 = v (we may assume that V # 0). So V is an irreducible R[A]-module. Thus, V does not have proper T-invariant subspace.

1104

Let A be an n x n matrix with entries in C. Show that A has n distinct eigenvalues in C if and only if A commutes with no nonzero nilpotent matrix.

Solution.

Then there exists an invertible n x n matrix P such that

(Inddana)

Necessity. Suppose that A has n distinct eigenvalues A', Xa,. . . , A, in C.

PAP-' = diag{ A 1 , . . . , An}.

If A commutes with some nilpotent matrix B , we have to show B = 0. Since A l l Az,. . , A, are distinct and

PAP-' = diag{X1,...,Xn)

commutes with P-'BP, P - l B P is a diagonal matrix. But the nilpontency of B implies that P ' B P is nilpotent. Hence we have P-'BP = 0. So B = 0.

Sufficiency. Suppose that the characteristic polynomial of A has multiple roots. We have to show that A commutes with some nonzero nilpotent matrix. Let diag(J1, Jz,.. . , J t ) be the Jordan canonical form of A and

PAP-' = diag(J1, Jz , . . . , Jt), where P is an n x n invertible matrix and J, is a Jordan block of order ei. Without loss of generality, we may assume that el > 1 (If all the ei = 1, then it is easy to see that A commutes with some nonzero nilpotent matrix).

Let B1 be the Jordan block of order e l , with 0 on the diagonal. Then JIB' = BIJl and Bl(# 0 ) is nilpotent. Let

B' = diag(B1, Bz,. . . , B,)

where Bi(i 2 2) = 0 E Me; (C). Then

B' . PAP-' = P A P - l . B'.

Taking B = P- lB 'P , we have B # 0, which is nilpotent, and AB = B A .

6

1105

Suppose V is a finite dimensional vector space over a field K , T : V -+ V a linear map such that the minimal polynomial of T coincides with the characteristic polynomial, which is the square of an irreducible polynomial in K [ T ] . Show that if %',T and 3 are any three non-zero vectors in V , then at least two of the three subspaces spaned by the sets (T*-?it}z>~, - {Ta?},20 and { T Z 3 } r > 0 - coincide.

Solution. V can be viewed as a module over the polynomial ring F[X] simply by

f(X).z = f(T).(z) for any z E V , f ( A ) 6 F[X]. Let {U~,U~,...,U,} be a base of V over F , A = ( u ~ ~ ) , ~ ~ be the matrix of T relative to the base. In general, a normal form for X I - A in M,(F[X]) has the form

(Stanford)

diag{l,. . . , l ,dl(X), . . + ,d,(X)}

where the d,(X) are monic of positive degree and d,(X)(d,(X) if i 5 j. By the structure theory of finite generated modules over P.I.D., there exist z,(z = 1 ,2 , ..., s ) E V such that V = F [ X ] . z l $ F [ X ] . z z ~ . . . ~ F [ X ] . z , where Ann(z,) = (d*(X)) . Here, according to the assumptions, the minimal polynomial m(X) of T is det(X1 - A ) , so

m(X) = d,(X) = det(X1- A ) .

Hence s = 1 and v = F[X] f z1 E?! F[X] / (m(X) )

is cyclic. Since m(X) is the square of some irreducible polynomial, V = F[X].z l has exactly two non-zero submodules. Obviously, the three subspaces generated by the sets {T'Z'},>o, { T E T } I > ~ - and {T*8} ,>o are non-zero submodules of V over F[X]. So at least two of them coincide.

1106

Let V be a finite dimensional vector space over with basis {wl,. . . , vn}. Let CT be a permutation on {q, . . . , v,} and thus induce a linear transformation A on V . Show that A is diagonalizable.

(HU7Wrd)

7

Solution. By re-ordering the elements 211, 'u2, * . , 'u,, we assume that

when u is expressed as the product of disjoint cycles (This decomposition may have 1-cycles). Let Wj be the subspace of V generated by {Kj-l + 1,. . , vij} for j = 1,2, - . - , s + 1 (io = 0, is+l = n). Then the Wj are invariant subspaces of A and V = W1 @ W2 @ .. . @ Ws+l. Let Mj be the matrix of Alwj : Wi -+ Wj relative to the base { ~ i ~ - ~ + 1 , . . - , v i ~ } of Wj overt. Then M = diag{M1,...,Ms+l} is the matrix of A relative to the base ( V ~ , V ~ , ~ ~ ~ , V , } . So it suffices to prove that every Mi is diagonalizable.

Hence, without loss of generality, we may assume that CJ is the n-cycle (v1,w2,..-,wn). The matrix of A relative to the base { V ~ , V ~ , ~ ~ . , V , ) is

M = 0 1 0 0 1 . 0 1 0 . . . ...

0 0 1 ... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 0 0 * . .

It is easy to see that the minimal polynomial of M is A" - 1, and thus M is diagonaliz able.

This completes the proof that A is diagonalizable.

1107

Let V be a finite dimensional vector space over the field of rational numbers. Suppose T is a non-singular linear transformation of V such that T-' = T2+T. Prove that 3 divides the dimension of V, and prove that if dimV = 3, then all such T's are similar.

Solution. Since T-' = T2 + T, T is annihilated by the polynomial X3 + X 2 - 1.

Obviously, X3 + X2 - 1 is irreducible over the field Q of rational numbers. Thus X3 + X2 - 1 is the minimal polynomial m(X) of T.

Now let n be the dimension of V over Q, A be the matrix of T relative to

some base of V , diag{l,. .-, l ,dl(X),.--,d,(X)} be the normal form for X I - A

( H U 7 V d )

n-s A

8

where the &(A) are monk of positive degree and di (X) ld j (X) if i 5 j. By the irreduciblity of &(A) = m(X) = X3 + X2 - 1, we have

&(A) = &(A) = . . . = &(A) = X3 + X 2 - 1.

Since det(X1- A ) = d l (X) . . .d , (X) ,

3 . s = deg(det(XI - A ) ) = n.

Thus we have proved that 3 divides the dimension of V . If dimV = 3, then X I - A is equivalent to diag(1, 1, X3 + X 2 - 1). The

. It follows that all the rational canonical form for A (or 7') is

T ' s are similar when dimV = 3.

1108

Let Fq be a finite field with q = p" elements, where p is a prime. Let 11 : Fq -+ Fq be the Frobenius automorphism II(z) = z p . Prove that II considered as a linear map over Fp is diagonalizable if and only if n divides p" - 1. (Here is a misprint. It should be "n divides p - l".)

Solution. (Hurvard)

It is wellknown that Fp C Fq is a Galois extension with

Gal(Fq/Fp) = {1,11,112,...,IIn-1}.

By the Normal Base Theorem, there exists a u E Fp such that {u, II(u), I12(u), ... , II"-l(u)} is a base for Fq over Fp. When II is considered as a linear map of Fq over Fp, the matrix of 11 relative to the base {u, II(u), I12(u), * a , II"-'(u)} is

0 1 0 ...

M = I.:. .I. .;. 1 1 : .;.).

1 0 ...

The normal form for XE - M is diag{ 1 , . . . , 1, A" - 1) and the minimal polynomial m(X) of M is A" - 1 = det(XE - M ) .

Suppose that 11 is diagonalizable as a linear map over Fp. Then m(X) = A" - 1 has no multiple root, and all the roots of A" - 1 are in Fp. On the other

9

hand, all the root of An - 1 forms a subgroup of FJ = Fp\{O}. Thus n divides p - 1.

Conversely, if n divides p - 1, A" - 1 has no multiple root and

A" - 1 = (A - 1) . (A - ad)(. \ - a 2 d ) . . . ( A - a("-l)d)

where d = and a is the generator of the group F;. Hence M is similar to diag{ 1, ad, . . . , u ( ~ - ' ) ~ } in Mn(Fp). So IT is diagonalizable as a linear map over Fp.

1109

Let A(t) be a non-singular matrix r.Lli)se elements are differentiable functions of real variable t . Let A'(t) denote the matrix formed by the derivatives of the elements. Show that the derivative of the determinant det A satisfies

Then

and

10

= det A1 + det A2 + . . . + det A,.

Hence we have proved that

d -(det A) = det A . trace(A’ . A-’). dt

1110

Let V be the vector space of polynomials p ( z ) = a + bz + cz2 with real coefficients a, b, and c. Define an inner product on V by

(a) Find an orthonormal basis for V consisting of polynomials 4o(z), 41(z),

(b) Use the answer to (a) to find the second degree polynomial that solves and q52(z), having degree 0 ,1 , and 2, respectively.

the minimization problem

(Courant Inst.)

11

Solution.

an orthonormal basis (a) Since 1, x, x 2 is a base of V , we can orthonormalization 1, x, x 2 to get

of V with degree 0, 1 and 2 respectively as usual. (b) Let

and Q O ( Z ) = z3 -Po(.)

such that ~ o ( z ) is orthogonal to V , that is

( x 3 - p o ( z ) , 4i(.)) = 0 (i = 0,1,2) .

Then for any p(x) E V ,

r l

3 (. , $a(z:)) = ( p o ( z ) , $i(.)) = Cl (i = 0,1 ,2) .

By an easy calculation, we get co = 0, c1 = 9 and cg = 0. Hence P O ( % ) = $z and no(.) = z3 - Zz. Obviously,

[:ix3 - 5z)2dx 3 = -. 8 175

12

Thus, when p ( z ) = gz,

is minimal, and

1111

Suppose that A is an n x n matrix and that

2 = ( :), Y = ( ;), y * = ( y 1 , * * . , yn). X n Yn

Suppose that all the entries of A, z, and y are real. (a) Show that there exist numbers a and b so that

det(A + m y * ) = u + bs.

(b) Show that if det(A) # 0 then a = det(A) and b = det(A) 3 y*A-lz. ( c ) Is it true that a = 0 if det(A) = O?

(Courant Inst.) Solution.

(a) Directly,

det(A + szy*)

a11 + SXIYI . * * a l n + Sz1Yn ... . . . ... an1 + SznY1 . * . ann + SXnYn

...

= det

= det U n 1 . . . a n n

Z l Y l * . . X1Yn [ det [ a21 ” ’ a2n + d e t . . . . . . . . . +s . anl . . . a n n

13

Hence det(A + sxy*) = a + bs for some a and b, for any s.

and (b) Since for any s , det(A + sxy*) = a + bs as in (a), we have a = det(A)

. . . . . . ...

b = det + det . . . . . . . . . . . . . . . . . . ... ann anl * . . a n n

. . . + . - . + det

an-11 a n - In

... X n Yn n n n

where A,, is the algebraic cofactor of ai j . If det A # 0,

All . . . An1 1 det(A)

A-l ___ ( . . . . . . . . . ) . Aln Ann . . .

Thus

All . * * An1 . . . . . . . . . ) . .

Aln . * . Ann = y*(det(A) . A-')Z

= det(A) . y*A-lx.

(c) If det(A) = 0, a = det(A) = 0.

14

1112

Let A be an n x n real matrix with distinct (possibly complex) eigenvalues, X I , Xz , . . . , A,, and corresponding eigenvector V ~ , V Z , . . a , v,. Assume that X I = 1 and that J A j l < 1 for 2 5 j 5 n. Prove that lirn Anv exists. Define T : C" + an by T ( v ) = lirn Anv. Find the dimensions of the kernel and image of T and give basis for both.

Solution.

matrix in illn(@) and P - I A P = diag{A1,A2,...,Xn} .

n-oo

11-00

(Courant Inst.)

Let P = (v l , vz , . . . , vn ) . Since A1,Az,...,Xn aredistinct, Pisaninvert ible

For any v E a", lim A'lw = lim P d i a g { X ; , X ~ , . . . , X ~ } . P - l . v

n-cc 12-00

= P . diag{l,O, ... , 0} . P-' . w

(Since lim (X,Y,) = lirn X , . lim Yn). Let (el, e2,. . . , en) be the standard orthonormal basis of (En. Then the matrix of T with respect to this basis is P ' - ldiag(l ,O, . . - ,O) .P ' . Let

n-cr, n+w n-cc

fn e n

Then ( f l , f z , . . . , f,) is a basis o f a n and

diag(l ,O,. . . ,O) = P'-P'-l -diag(l ,O,. . . ,O)P'P'-l

is the matrix of T with respect t o the basis ( f l , f z , . . . , f n ) . Hence {fl} is a basis of Im(T), {f2,...,fn} is a basis of ker(T), and dim(Im(T)) = 1, dim(ker(T)) = n - 1.

1113

Let A be a matrix. Define 1 1 3! 5 !

sin(A) = A - -A3 + -A5 - * .

For

express sin(A) in closed form.

15


Denote B = ( -3 - " ) 7 .

Then B is similar to

and

1 -1 0 10 1 -1 0 bviously,

( ; '1 )-'= ( 1 ) . Hence

sin(A)

- ( 1 -1 ).[;.(4 0 10 0 ) - L ( T ) 3 . ( 4 3 3! 4 o 103 O ) -

1 1

where

16

1114

Let A be the 9 x 9 tridiagonal matrix

A =

' - 2 1 1 -2 1

1 . a . - 0 .

* * . -2 1 1 -2

All entries not shown are zero. (a) Show that

(b) Use part (a) to show that A has all negative eigenvalues. (c) Let B be the following 10 x 10 tridiagonal matrix, which agrees with A

except in the first row and column:

B =

1 -3 -3 -2 1

1 -2 1

1 * . * .

Let A,,(B) be the largest eigenvalue of B . Show that A,,,(B) > 1. (d) Show that B has 9 negative eigenvalues.


(a) A direct verification shows that (a) is true. (b) Since zTAz 5 0 for all z E lR9 and z T A z = 0 if and only if z = 0, all

the eigenvalues of A are negative. (c) Obviously, the symmetric matrix X,,,I - B is semi-definite positive.

Then for any Y E Elo, YT(X,,,I - B)Y >_ 0, this is, Y T B Y 5 X,,,YTY. Taking YT = (1 , -1 ,0 , - - . , 0 ) , we have Y T B Y = 5 5 X,,,.2. Thus A,, > 1.

(d) Let Vl = { (O ,z1 , . . . ,~9 ) I zi E IR} 5 Elo. Then for any Y E V1, Y T B Y 5 0 and Y T B Y = 0 if and only if Y = 0. If B has more than two non- negative eigenvalues. Let A1 and A 2 be two of them, we have y1, y2 E R I O such

17

T that y1ly2, y1 y1 = yTy2 = 1 and Byi = Aiyi (i = 1 , 2 ) . Let Vz = (yl,y2), the subspace spaned by y1 and y2. Then for any y = ulyl + uzy2 E V2, yTBy = a?A1 + u; * A 2 2 0.

Since dim V1+ dim Vz = 9 + 2 = 11 > 10,

there exists 0 # y E V1 n VZ. Then we have yTBy < 0 (y # 0, y E V1) and yTBy 2 0 (y E VZ), a contradiction. Thus B has 9 negative eigenvalues.

1115

Let T be a real symmetric, positive-definite, n x n matrix with distinct eigenvalues A 1 > A2 > . . . > A,. Show that

I T X I v 2EV-{O} 1x1 ' A 2 = max min -

where V ranges over all two dimensional subspaces of Bn and 1x1 is the Eu- clidean norm of 2 .

Hint. Show = by showing 5 and 2 separately. You may wish to express T in a basis of eigenvectors.


By assumption, there exists an n x n orthogonal matrix P such that P'TP = diag(A1,A2,-..,Xn). Since for any orthogonal matrix H and any y E B", (Hyl = Iy1, we have

IP'-ldiag(Xl, Az,. . . , A,) . P-lxI min lTxl - min - -

2EV-10) 1x1 Z€V--IO) 1x1 - Idiag(Al,...,A,). P-lxI - min

2EV-IO) IP-1x1

If V is a 2-dimensional subspace, P - l V is also two-dimensional. Hence

Idiag(A1, . . . ,A,) .x1 max min w = m a x min v Z€V-{O} 1x1 v Z€V-{O) 1x1

when V ranges over all two dimensional subspaces of lR".

V = (e1,eZ). Then Now let {el, e 2 , . . . , en} be the standard orthonormal basis of Rn, and

Idiag(Al,-.., L ) z l min x €V - to1 1x1

18

= A 2 (A, > A 2 > 0).

It follows that

2 Az. I T X I

v Z€V-{O) 1x1 max min -

On the other hand, for any two dimensional subspace V of 1R". we can find an orthogonal basis {fl, fi,.. . , fn} of IR" such that V = (fl, fz). Let

(fl, fz , . . . , fn IT = Q(e1, ez,. . . , e, lT. Then Q = ( q i j ) , x n is an orthogonal matrix, and if 0 # x = alf1 + a z f z E V .

Idiag(A1,.-., A,) - X I 1x1

so.

and

Thus

when V ranges over all two dimensional subspace of B".

19

1116

For any n x n matrix P , consider the sum 00

R(P) = c Pk k=O

M

(a) Prove that if JIPklJJ < 00, then (I-P)-' exist and R(P) = ( I -P)- ' .

(b) Assume that IlPll < 1 in a matrix norm induced by a vector norm.

(c) Use part (a) to compute the inverse of

1 1 0 0

k=O

Prove that ll(I - P)-'ll 5 *.

A = [ ! i ; r ) . (Courant Inst.)

Solution.

A', Az,. . , A, be the n eigenvalues of P and (a) First, we claim that the norms of all eigenvalues of P are < 1. Let

A1 * Q-lpQ= ( e f . )

be the Jordan form of P. Denote

rjm(z) = 1 + z + * * . + z".

sm = E + P + P2 + . . . + P"

Then ~ m ( A l ) , r j m ( X ~ ) , ~ ~ ~ , ~ m ( X , ) are the n eigenvalues of

00

Since C IJPk(l < co and the norms over vector space are all equivalent, k=O

00

k=O

20

is convergent. Since

1 Q , lim Q - ~ s , Q = Q - ~ ( lim S,

m-w m-w

W

lim dm(X,) exists, that is, C is convergent (i = 1,2 , . . - ,n ) . Hence m=O m-oo

pal < l ( 1 5 i 5 n). Thus all the eigenvalues of I - P are non-zero, and (I - P ) is invertible.

Since

S , ( I - P ) = I -P"+' ,

lim Sm = lim ( I - P"+')(I - PI-' rn-00 m-+m

I - lim ~ + l ) ( I - PI-' = ( n-rn

= (I - O)(I - P)-' = ( I - P ) - ' .

Thus 00

R ( P ) = Pk = ( I - I J - 1 .

k=O

(b) By definition,

l l ~ l l = sup { #lv non-zero vector . I Since llPll < 1, for any non-zero vector u, llPull 5 llPll . llull < llull for the corresponding vector norm.

Now suppose ( I - P).u = 0 for some vector u. Then llull = IlI-uII = IIPuII. We must have u = 0. The matrix I - P is therefore invertible, and we can write ( I - P)-' = I + P(I - P)-'. Then we have

21

Then A = I - N . So

1117

Let G be a pgroup and G x V + V be a linear action of G on a finite vector space over a finite field Fan. Using Sylow theory, prove that there exist a basis of V in which all the transformation v + a . v(a E G) are unipotent matrices.

Solution. Let

( CoZumbia)

G' = {V + V, w --+ 0 * 01g E G}.

Then G' is a pgroup, since there is a surjective homomorphism G -+ G' ( 5 EndF(V)). We fix a base of V over Fan. Then G' is isomorphic to a subgroup H of GL(m, Fan), where m = dirnp,, V .

It is well known that

m

j=1

and

U =

is a Sylow psubgroup of GL(m, F p n ) . By Sylow's Theorem, there exists some P E GL,(m,Fpn) such that PHP-' U . Thus, if we change the base of V via P , the corresponding matrices of the linear transformations in G' are in U, which are unipotent matrices.

22

1118

Let S be an endomorphism of a finite dimensional vector space V over a field F whose characteristic polynomial is not equal to its minimal polynomial. Show that there is an endomorphism T of V so that T commutes with S but T is not a polynomial in S (T is a polynomial in S if T = aol+alS+a2S2+. . .+akSk for some k and ai E F. )

Solution. Let

(Stunfod)

F [ S ] = {f(S) : V --t V I f(X) E F[X]} .

Then as F-vector space, dimF F [ S ] = degrn(X), where r n ( A ) is the minimal polynomial of S. Since the minimal polynomial of S is not equal to its characteristic polynomial, dimF F [ S ] < dimF v.

On the other hand, let d l ( X ) , &(A), .. . , & ( A ) be the invariant factors # 1 of S and let ni = degdi(X), then by Frobenius theorem, the dimension of the vector spcae over F of matrices commutative with the matrix of S is

S

N = C(2.s - j=1

2 j + 1)nj.

0 bviously, 8

N 2 C nj = dimF V. j=1

So there is a linear transformation T of V such that T commutes with S but T is not a polynomial in S.

1119

Let A be an n x n real matrix, all of whose (complex) eigenvalues are real and positive. Show that for any integer rn 2 1, there exists a t least one n x n real matrix B with B" = A.

Hint. Make use of the Jordan form S + N of a conjugate of A, where S is diagonal and N is strictly triangular.

(Stanford)

23

Solution. Suppose first that

A =

be a Jordan block with a real and positive. For any integer m 2 1, let b = m&

and

B =

b 1 0 O b 1

0 0 * . - b 1 0 0 . . * O b

It is easy to see that

I * b" Chb"-l Cibm-1 . . . C;-lbm-n+l 0 b" cibm-1 . . , ~;-2bm-n+2

B m = :

'.. chbm-l ... b" ... ... 0

A = Q-lB"Q = (Q-lBQ)"

L Obviously, XE-A and XE-B" have the same invariant divisors (1,. . . , 1, (A- a).}, A and Bm are similar. Thus

for some invertible n x n real matrix Q.

positive. Then there exists an n x n invertible real matrix P such that Now let A be an n x n real matrix, all of whose eigenvalues are real and

F I A P = diag(J1, J2, . , J k )

where Ji (1 5 i 5 k) are Jordan blocks with diagonal elements real and positive. As proved in the above, for any integer m 2 1, any 1 5 i 5 k, there exists real matrix B, such that By = Ji (1 5 i 5 k). Hence

P-lAP = (diag(B1,B2,. . . ,Bk))m

24

and

where

B1 0 B2

0 Bk

m

P-'] = Bm,

1120

Let V be a finite dimensional vector space over an algebraically closed field K , and let T : V 4 V be a linear transformation.

(a) Show that there are 2'-invariant subspaces V , V and elements of a; E K such that V is the direct sum of the V, and (T - ail) : V, + V, is nilpotent. (A transformation N is nilpotent if N" = 0, for some n 2 1.)

(b) show that there are polynomials S ( T ) , N ( T ) E K[T] such that T = S(T)+N(T) , S ( T ) : V -+ V is diagonalizable, and N ( T ) : V -+ V is nilpotent.

Hint. Use the Chinese Remainder Theorem. ( Stanford)

al)ell, * . . , (X-al)el'l ; (X-a2)e=, * * * , ( A - a Z ) e z T . , ; (A-a,)enl, . . ., (A-an)e-

Solution. Viewing V as a module over the polynomial ring K[A] via T. Let (A -

be the elementary divisors of VqX], where a1, a2,. . . , an are distinct and

By the structure theorem of finitely generated modules over PID, we may write

where all the w,j E V and

A n n ( ~ ; j ) = ((A - ~ ~ i ) ~ ~ j ) .

Denote V , j = K[A]wij. Then

25

The K j ' s are T-invariant subspace and (T-a i l ) : V;:j -+ vj is nilpotent. Thus (a) is proved.

Since ( A - al)e*'l , (A - a 2 ) e 2 r 2 , . . . ( A -

are pairwise relatively prime, there exists some S(X) E K[X] such that

S(X) ai( mod ( A - ai)eir; ), (1 5 i 5 n),

by Chinese Remainder Theorem. Then

S ( T ) . waj = Cya . waj

for all wij and it is easy to see that S ( T ) : V -+ V is diagonalizable. Let N ( X ) = X - S(X). Obviously, T = S ( T ) + N ( T ) and X - ai I N(X) for any 1 5 i 5 n. So the minimal polynomial m(X) I N(X)h for some positive integer h. Thus N ( T ) : V + V is nilpotent. Thus (b) is proved.

26

SECTION 2 GROUP THEORY

1201

Let G be the group of real 2 x 2 matrices, of determinant one. Describe the set of conjugacy classes of elements of G.

Solution.

A 1 - A2 = 1.

(Hamud)

Let g E G, A 1 and A2 be the eigenvalues of g viewed in M2(@). Then

i) If A1 = A 2 , then A 1 = A 2 = 1 or A 1 = A 2 = -1. g is similar to

in GL2(R).

In case a), g =

In case b) g =

In case c), let A be an invertible 2 x 2 matrix with real entries such that

N = ( - d ) - ? ( -1 0 l ) A

if d = det A < 0, then N E G and

NgN-'= ( ; ; ) or

1 -1 NgN-'= ( ) .

27

( 1 1 1 ) or ( 1 -1 ) i n G . But ( A i ) isnot Hence g is conjugate to

conjugate to ( q' ) in G, for if

A * ( i : ) A - ' = ( ;') where A = ( E:: :tz ) E G, then we have a21 = 0, all = - a 2 2 and

de tA = -a!l = 1 which is a contradiction. Thus in case c), we have two

conjugacy classes [( i!i :)] and [( i!i . ')I* In case d), g is similar to ( 0' 21 ) in G L ~ ( B ) . AS in case c), g is

) or ( i1 1;) i n G a n d ( i1 '1 ) similar (conjugate) to ( 0' is not conjugate to ( i1 1; ) in G . Thus in case d) , we have two conjugacy

classes [ ( i1 yl')] and [ ( 0' 1; )] . ( 2 x", ) in ii) If X i # A2 and X i E B (i = 1,2) , then g is similar to

GL2(B). There exists M E G L z ( R ) such that

We take N = d-4 . M i f d = d e t M > 0 or

N = ( - d ) - i ( -1 0 ' ) M

if det M = d < 0. Then N E G and

Thus g is conjugate to ( x", ) in G . So in this case, we have the conjugacy

classes

28

iii) If XI # A2 and X i E C\E (i = 1,2), we denote A1 = cos8 + i s i n 8 and X2 = cos8 - i s i n 8 (since XI . A2 = 1 and trace ( g ) = A 1 + A 2 E E) , where

181 < T , 8 # 0. g is similar to ( cos8;isino O ) in G L ~ ( C ) , cos 8 - i sin 8

) in GL2(C). cos8 sin8 ( -s in8 cos8

and also similar to

As it is well known, that two matrices A , B E M 2 ( B ) are similar in M2(~,7 implies that A and B are similar in M2(nZ). So there exists M E G L 2 ( E ) such that

cos8 sin0 M g M - I =

As in the above, we conclude that g is conjugate to

cos$ sin8 -s in6 cos6

or

in G. But

and

cosB -sin$ sin8 C O S ~

cos8 sin8 -sin$ COSB

cos0 -s in8 sin0 cos8

are not conjugate in G, for if

such that cos8 sin8 ) A-l = ( cos8 -s in0

det A = -(a:l + ail) = 1,

which is a contradiction. So in this case, we have the conjugacy classes

29

and cos6 -s in6 sin6 cos6

To sum up, the set of conjugate classes of elements of G is

{ [ ( $ x", ) ] ~ A I , X z E El and XI 3 XZ = 1

" {[ ( i Y ) ] Ib=* l}

{ [ ( -s in6 c0s6 cos6 )] 10 < (61 < .) U

1202

Let G = GLn(Fq) , the group of invertible n x n matrices over the finite field Fq with q = p" , p a prime, U = Un(Fq), the subgroup of upper triangular matrices with 1's on the diagonal

a) Calculate the orders of G and U . Deduce that U is a Sylow psubgroup of G.

b) Deduce that every psubgroup of G is conjugate to a subgroup of U . c) Determine the number of G-conjugates of U. d) Show that g E G has ppower order iff g = I + N with N n = 0. e) Show that G contains elements of order qn - 1 Hint. Make use of Fqn.

(Columbia) Solution.

a) When forming a matrix in G L n ( F q ) , we may choose the first row in qn - 1 ways (a row of zeros not being allowed), the second row in qn - q ways (no multiple of the first row being allowed), the third row in qn - q2 ways (no linear combination of the first two rows being allowed), and so on. Thus we can conclude that the order of GL,(Fq) is

30

When forming a matrix in Un(Fq) , we may choose the first row in q n - l ways, the second row in qnP2 ways: the third rows in q n - 3 ways and so on. Hence the order of Un(Fq) is

SO U = Un(Fq) is a psubgroup of G = GLn(Fq) . Since p j [G : U ] = n n (P'~ - l ) , U is a Sylow psubgroup of G.

i = l b) I t follows directly from Sylow's Theorem. c) By Sylow's Theorem, the number of G-conjugates of U is equal to [G :

N G ( U ) ] where N G ( U ) is the normalizer of U in G.

n, there exists some Suppose g = ( g i j ) n x n E N G ( U ) , or gUg-' = U . Then for any 1 5 i < j 5

U l n 1 U12 . . . ... 0 1 ~ 2 3 . . .

u=[.;. .I. .;. ::: ... j E V

such that g(I + Ej j ) = u g where Eij is the matrix with a lone 1 in the (i,j) place, 0's elsewhere. Checking the last row of g ( l + E i j ) and u g , we get gni = 0. So g n , = 0 if i < TI. Then checking the (n - 1)th row, we get g n - l , j = 0 if i < n - 1. By this way, we get g i j = 0 if i > j. Hence g E T = Tn(Fq) , the set of matrices (ai j ) E G with a;j = 0 (0 5 j < i 5 n).

Conversely, if g E T: It is easy to see that g U g - ' & U (by matrix multiplication). So g N G ( U ) and we have N G ( U ) = T .

n b

We can define a map 19 : T + F; x ... x F; by mapping a matrix onto its principal diagonal, matrix multiplication shows that 0 is a group epimorphism whose kernel is precisely U = Un(Fq) . So the order of T is

n

i = l Hence the number of G-conjugates of U is n ( q i - l ) / ( q - l)n.

d) Suppose g E G has ppower order. Then ( 9 ) is a psubgroup and ( 9 ) is conjugate to a subgroup of U by b). There exists an h E G such that h g h - l E U . Denote h g h - l = I + M where M is a matrix with zero on and below the diagonal. Obviously Mn = 0 and g = I + h - l M h = I + N where N = h - l M h and N" = h- lM"h = 0.

31

[GI = U gHg-' IgEG

Conversely, if g = I + N with N" = 0, the Jordan canonical form for N is a matrix with zero on and below the diagenal. Thus g is similar (conjugate) to some matrix in U . So the order of g is ppower.

e) Consider the finite field Fqn as an extension of the field Fq and as a vector space over Fq. For any a(# 0) E Fqn, we have a linear map al : Fqn --f FqR, a l ( z ) = az (over Fq) which is invertible. We fix a base of Fqn over Fq, then we can obtain a group homomorphism (T : FLf. --$ G = Gn(Fq), ~ ( u ) = at. Obviously, (T is injective. It is well known that F;,, has an element with order q" - 1. It follows that G contains element of order q" - 1.

5 [G : N ( H ) ] . (IHI - 1) + 1.

1203

32

(b) First we claim that G is cyclic. Suppose that G is not cyclic. Then for any a E G , (a) < G. So (a) is contained in some maximal subgroup of G (since G is finite). Let H be a maximal subgroup of G. Then we have G = U gHg-l

U E G by assumption. By (a) G = H . This contradicts the maximality o f k . Hence G is cyclic. Now, it is easy to see that G is cyclic of prime power order since all of its maximal subgroup are conjugate.

1204

Suppose that H and K are normal subgroups of a finite group G and that

(a) Give an example to show that G / K need not be isomorphic to H . (b) Show that if H is simple, then G I K N H .

G I H = K .


(a) Let G = Q8 = { f l , & i , f j , 3 t k } be the Hamilton’s Quaternion group and H = (i) = {3tl,*i}, K = (-1) = {*l} be the subgroups of G generated by i and -1 respectively. Obviously, we have H a G, K a G and GIH is cyclic of order 2, so it is isomorphic to K . But G I K = { 1, i, j , I c } N K4 (Klein four group), K is not isomorphic to H .

(b) Let (1) = KO a K1 4. .. a K,-1 u K , = K be a composition series of K . Since GIH N K , there exist subgroups Hi such that Hi I> H , H i I H N Ki ( i = O,l , . . . ,n ) and Hi ~ H i + l (i = O,- . . , n - 1).

_ - _ -

Suppose H is simple, then

(1) u H a H 1 a ... a H,-l a H , = G

is a composition series of G with composition factors H and Hi /Hi - l N

Ki/Ki- l (i = 1,2, * * * , n). On the other hand,

(1) = KO a K 1 a . . . a K n P 1 a K U G

is a normal series of G with factors KiIKi-1 (i = 1,2 , . . . ,n ) and G / K . By the Jordan-Holder Theorem, we have the isomorphism G / K N H .

1205

Prove that if G is a finitely generated infinite group then for each positive integer n, G has only finitely many subgroups of index n.

33

Hint. Let H 5 G and define HG = g-lHg. Show that if H 5 G

(finite index) then there exists a homomorphism of G/HG into S[G:XI, where S, denotes the symmetric group on n letters.

Solution. Let H be any subgroup of G such that [G : H ] = n. Then G x G I H + G / H ,

g . ( z H ) = (gx ) . H defines an action of G on G I H , where GIH denotes the set of all left cosets z H ( z E G) . The kernel of this action is HG = n g-lHg.

Hence the group GIHG is isomorphic to a subgroup of S, , the symmetric group on n letters. This induces a homomorphism G + S, with HG as its kernel.

Since G is finite generated, there are only finite number of homomorphisms from G to S,. It follows that the set

g € G

( Co2umbia)

g € G

{HGIH < G , [G : H ] = n}

is finite. Since G/HG is finite, G/HG has only finitely many subgroups of index n. Thus G has only finitely many subgroups of index n.

1206

Let

G = { ( a!l ) la E lR*, b E lR C_ GL(2,lR). 1 Show that G is solvable but not nilpotent.

Solution. (Columbia)

a!l ) to its diagonal Let 0 be the map G + R* x lR* mapping ( ( a , ~ ' ) . Matrix multiplication shows that 0 is a homomorphism, and

ker8= { ( ) 16tnJ- Obviously, N = ker0 (21 (lR,+)) is an abelian subgroup of G, and GIN N

E x IR* is also abelian. Hence G is a solvable group. Suppose

34

the center of G. Then for any ( ; x!!l ) E G ,

So ay+ bx-' = xb+ya-l for any x # 0, y E IR. It follows that b = 0, a = f l . Thus

A direct discussion as above shows that C(G/C(G)) = ( 1 ) . Hence

C1(G) = C2(G) = * * - = C,(G) = a * * =$ G

where Ci+l(G)/C,(G) = C(G/Ci(G)) (Co(G) = 1) . So G is not nilpotent.

1207

Let p be a prime and let V be an n-dimensional vector space over Fp. Let G = G L F ~ ( V ) . Recall that IGI = (p" - l ) ( p n - p )

(a) Recall that a linear transformation is called semisimple if its minimal polynomial is separable. Prove that a transformation T E G is semisimple if and only if Tprn-l = 1 for some positive integer m.

(b) Let H be a subgroup of G of order a power of p . Show that H can be simultaneously upper triangularized, that is, there is a basis of V with respect to which all of the elements of H are upper triangular.

- (p" - p"- l ) .

Hint. Find a Sylow psubgroup of G. (Indiana)

Solution. (a) Let T E G and f ( X ) be the minimal polynomial of T over Fp. Then X

1 f ( X ) . If T is semisimple, that is, f(X) is separable (remark: here separability means that f(X) has no multiple roots), then f ( X ) = f l ( X ) f 2 ( X ) - - . f k ( X ) for some distinct irreducible polynomials fi ( 1 5 i 5 k ) . Denote ni = degfi(X) and m = n1 . n2 'nk. Let E be a splitting subfield of XPm - X over Fp and Ei = Fp[X]/(fi(X)) ( 1 5 i 5 k ) . Then IEl = p" and Ei is a field of order pni. Since nilm, E, is isomorphic to a subfield of E. So E contains an element ~i whose minimal polynomial over Fp is f i ( x ) . Since ~i # 0 and 7.frn-l = 1, f i ( A ) ~ ( A P r n - l - l ) , ( 1 5 i 5 k ) . Hence f(X) = fl(X).fi(X)...fi~(X)I(X~~-~-l) . Thus Tprn-l = 1.

35

Next suppose TPm-' = 1 for some positive integer m. Then f ( X ) I ( P " ' - ' - 1). Since (XP"'-' - 1)' = -XP - 2 # 0, f ( X ) has no multiple roots. Thus T is semisimple.

(b) Let U = U,,(V), the subgroup of upper triangular matrices with 1's on the diagonal. Then

m

JUJ = p"-l . p " - 2 . . . p = pv (when forming a matrix in U, we may choose the first row in pn-' ways, the second row in pn-2 ways and so on). Since

)GI = (p" - l)(pn - p ) * * * (p" - p"-') n n-1

= p- . (p" - 1) . (p"-' - 1). . . ( p - I),

U is a Sylow psubgroup of G. Now let H be a subgroup of G of order a power of p. By Sylow's Theorem,

H is contained in some Sylow psubgroup K of G and K is conjugate to U . Hence there exists an element g in G such that gHg- l 5 gKg-' = U. Thus we have proved that H can be simultaneously upper triangularized.

1208

Let p and q be distinct prime numbers. Let G be a group of order p 3 . q such that its commutator subgroup K is of order q. Let H be a pSylow subgroup of G.

(a) Show that H is abelian and G = H K . (b) Show that there are elements h E H and k 6 K such that hk # kh. (c) From (b) show that p divides q - 1.

(Indiana) Solution.

(a) Since K is the commutator subgroup of G. K is normal in G and G / K is abelian. So H - K = K H is a subgroup of G. Since IK( = q , IHJ = p3, H n K = { e } and lHKl = p3 q = IGI. Hence G = H K and H 21 H / ( e ) = H / H n K 21 H K I K E G / K is abelian.

(b) Suppose that for any elements h E H and k E K holds hk = kh. Since K is abelian and H is abelian, G = H K is abelian. This contradicts the fact that the commutator subgroup K of G is of order q. This proves (b).

( c ) First we claim that H is not normal in G. Otherwise, for any h E H and k E K , hkh-lk-l E H n K = { e } and so hk = kh. By Sylow Theorem,

36

the number of pSylow subgroups of G is greater than 1 and divides q . So it is q , Again by Sylow Theorem, plq - 1.

1200

(a) Let p , q be primes, p > q > 2 . Let G be a group of order pq’. Show G

(b) What can you say if q = 2 (and p > q)? has a subgroup of order p q .

(Indiana) Solution.

(a) Let rp be the number of Sylow psubgroups of G. Then, by Sylow’s Theorem, rP lqz and plrp - 1. Since p , q are primes and p > q > 2 , it is easy to see that rp = 1. So G has only one Sylow psubgroup H , which is normal in G and of order p. By Cauchy’s Theorem, G also contains an element of order q . So G has a subgroup K of order q. Thus H . K is a subgroup of order p . q since H is normal in G.

(b) If q = 2, G may not contain a subgroup of order p . q . For example, Aq, the alternating group of degree 4, is a group of order 3 . 2 2 . A4 does not have a subgroup of order 6 .

1210

Let A be the abelian group on generators el f, and g, subject to the relations

9 e + 3 f + 6 g = 0, 3e + 3f = 0, 3 e - 3 f + 6 g = 0.

Give a decomposition of A as a direct sum of cyclic groups of prime order or infinite order.

Solution. Let F be the free abelian groupgel $Zez $Ze3, K be the subgroup of F

generated by f1 = 9el + 3ez + 6e3, fz = 3el + 3ez and f3 = 3el - 3e2 + 6e3. Obviously A E F / K . Denote

(Stanford)

9 3 6

37

It is easy t o get the normal form diag{ 3,6,0} of M in M3(Z) and to find

P = ( O 1 0 0 -1 1 )

0 1 1

and

Q = ( O 0 -1 1 -2 -1

such that Q M P = diag{3,6,0} ( P and Q are invertible matrices in M 3 ( Z ) ) . Let

(e ; , e i , e$>' = P - l ( e l , e2, e3)'

and (fi, fL f4)' = Q . (fi, f 2 , f3)'.

Then F =Zei $Ze', $ZeL

and K = Zfl + Z f 2 +Zf3 = XCe; @ Uel, .

so

A S F / K = 2 e i $Zek $57eL/Uei @ Saei

r" 57/(3) $2/(6) @Z 2 2 2 $ 2 3 $573 $2.

1211

Let M be an n x n matrix of integers. Suppose that M is invertible when viewed as a matrix of rational numbers, i.e., that there exists an n x n matrix N with rational entries so that M N equals the n x n identity matrix. View M as an endomorphism of 2".

a) Show that 2"/MZn is finite. b) Show that the order of Zn/M57" is equal to the absolute value of the

determinant of M . (Stanford)

Solution. a) It follows directly from b). I t is also obvious from the facts that the

map g : Zn/MZn + Z " / l M p " , 6 ( r ) = M * X is injective, where [MI is the determinant of M , M * is the adjoint of M , and the fact p n / l M p n l = ( I det MI)".

b) Let D = diag{dl,dz,... ,dn} be the normal form of M in M,(Z), where the di # 0 and dildi+l (i = 1 , 2 , . . . , n - l), and P , Q E M,(Z) be invertible matrices in M,(Z) such that D = Q M P . Obviously

det D = dl . d z . . . d, = det Q . det M . det P = det M or - det M .

Let {el, e2, . . . , e n } be a base for the free module Z". Denote

(ei ,el , , . . . ,ek)l = P-'(el ,ez, . . . ,e ,) ' .

Then {ei ,e l , , . . . ,ek} is another base ofZ". Let I /

(fl , fz,. .. , fn)' = M . (e l , e2, . , en) ' = MP (ei, el,, . . . , en) .

Then M P is generated by { f l l f 2 , . . . , f n } . Since Q is invertible in Mn(Z) , M Z n can be generated by {f;, f;, . . . , f;} where

(fil f; 7 * * * , f: )' = Q . (fi 1 f~r . * * f n )'.

0 bviously,

(f; 1 f;, . . , f; 1' = Q . (fll f~ , . . . , fn ) I

I / =

= Q M P - (e i , eh , . . . , e,)

diag{dl, dz, . . . , d, } . (ei, eh, . - . , en) . I I

Hence

1212

Let D =a[ [ ] with E = *. Calculate the order of the additive group G = D2/K where K is the D-submodule of D2 generated by ( 2 t + 11[ - l), ([ + 2 , ( - 4) and (21,21). Then express G as a product of cyclic groups.

(Haward)

39

Solution.

(as additive group). Since ( is a root of the irreducible polynomial z2 + z+ 1, D = Z[ t ] = Z @Z<

Let {elle2} be a base of the free D-module 0’. Then

0’ = Del @ Dez = Z e l @Z(el @Zez @Z<e2,

{ell(el,e2,(e2} i s a baseofD2asZ-module, and {(1+2<)e1+(-1+<)ez1(2+ ()el + (-4 + () . e2, 21. el + 21. e2} is a generating subset of the D-submodule K of 0’. For any a + bt E D ,

(a + b t ) [ ( l + + (-1 + t ) e ~ l = a ( e l + 2tel - e2 + (e2) + b(-2e1 - (el - e2 - 2(e2),

(a + b0[(2 + ()el + (-4 + Oezl a(2el +<e l - 4ez + (ez) + b(-el+ (el - e2 - 5(ez),

(a + b()(2le1+ 21e2) a(21 . el + 0 . <el + 21 - e2 + 0 - <ez)

+b(O - e l + 21 (el + 0 . e2 + 21 . (e2).

- -

=

It is easy to see that {el + 2te1- e2 + (ez, -2el- (el - e2 - 2(e2,2el +(el - 4e2 + 5e2, -el + (el - e2 - 5(e2, 21el+ 21e2,21(el + 21(e2} is a generating subset of K as Z-module (additive group).

Denote 1 2 - 1 1

-2 -1 -1 -2

A = 1 ;; ,9 ~ 5 ~ .

N = [ ; ; 0 0 0 ; ;l] 0

0 21 0 21

It is routine to get the normal form

1 0 0 0

for A in hf6x4(Z) and to find invertible matrices P E &(Z) and Q E 1M4(Z) such that PA& = N . It follows that D 2 / K -.2/(3) @Z/(21) and the order of D 2 / K is 63.

40

1213

1 rtl Prove that SL(2,Z) is generated by ( ) and ( i1 ) where

SL(2 ,Z) is the group of 2x 2 matrices with integral coefficients and determinant = 1.

Solution. (Stanford)

Suppose that M =

If both a and c are nonzero, we claim that ( 1 ) is a prodct of the

matrices in SL(2 ,Z) with the form

for some b',c',d' EZ.

that if a = 1, Since det M = ad - be = 1, a and c are relatively prime. It is easy to know

and if c = 1,

If a > c > 1, there exist some q , a1 E Z such that 1 5 al < c and a = qc+al. Hence

for some b' E Z. Similarly, if c > a > 1, let c = ha + e l , 1 5 c1 < a, then

for some d' EZ.

41

According to the above discussions, it can be derived that M =

is a product of the matrices in SL(2 ,X) with the form

for some b', c', d' E Z. So to prove that SL(2 ,Z) =

1 3zl 0 -1 ) and ( generated by ( ), i t suffices to prove that all

are in (( 1 fl cc) , ( i1 )) where b ' , c ' , d 'EZ .

It is easy to check this by the property of the elementary matrices. This completes the proof.

1214

Let G be a finite group, K a normal subgroup of G and P a Sylow subgroup of K . If N is the normalizer of P in G (i.e., N = {g E G 1 gPg-' = P } then show that G = K . N .

Solution. For any g E G, we have 9Pg-l C gKg-' 2 K since K is normal in G.

Hence gPg-l is also a Sylow subgroup of K . By Sylow Theorem, there exists h E K such that gPg-l = hPh-l . Hence h-'gP(h-'g)-' = P and h-'g E N . So we have g = h . h-'g E K - N . Thus we have G = K . N .

( I n d i a n a )

1215

Let P be a Sylow subgroup of a finite group G. Let N = { x E G I xPx-' = P } . Let H be asubgroup of G, H 2 N . Prove: If y E G such that yHy- l = H , then y E N.

( I n d i a n a )

42

Solution. Obviously, P is a Sylow subgroup of H . Since yPy-' & yHy-' H ,

yPy- ' is also a Sylow subgroup of H . So there exists an h E H such that yPy-' = hPh-'. Hence h-'yP 1 (h-ly)-' = P and h-'y N . So

y = h . h - l y E H . N c H .

1216

Let G be a finite group. The probability of G to be commutative is defined by

I{(a,b) E G x G I ab = ba}J P ( G ) =

IG x GI (a) If G is not commutative, prove that P ( G ) 6 5/8. (b) Give an example such that P(G) is exactly 5/8.

(Stanfod) Solution.

(a) Let C(G) be the center of G and CG(U) = { b E G I ab = ba} be the centralizer of a in G for any a E G. If G is not commutative, then, obviously, IG/C(G)I L 4.

Since

J{(a,b) E G x G I ab = bu}l =

P(G) =

L

1 1 5 2 4 8 - . (1 + -) = -

43

(b) Let G be the dihedral group 0 4 = {z*y j I 0 5 i 5 1 , 0 5 j 5 3,z2 = y4 = 1, zy = y3z}. Then IGJ = 8 and C(G) = (1, y}. For any a E G - C ( G ) , obviously, 3 5 l C ~ ( a ) l < 8, and so ~ C G ( U ) / = 4. It is easy to see that P ( G ) = 518.

44

SECTION 3 RING THEORY

1301

a) Prove the ring Z [ n ] is Euclidean. b) Using a), find all integer solutions to the equation y2 + 2 = x3.

(Hurvard) Solution.

a) It is readily known that Z [ n ] is a subring of a', hence an integral domain. For any a = m + nz/--2 E Z[z/--2], we define 6 ( a ) = m2 + 2n2. So we have a map 6 :Z[-]* + Z?, a H 6(a) .

Suppose that a, b # 0 E a[-]. Then ab-' = p + v.\/---z where p and Y are rational numbers. We can find integers u and v such that Iu - pl 5 1/2 and lw - vI 5 1/2. Then

a = b ( p + v G ) = b[(u+p - u) + (v + v - v)Q]

= b q + r

where Q = u + v Q is in Z[-] and

r = a - bq = b(p - u) + b(v - v)Q.

Obviously T E Z[-] and

6 ( T ) = 6 ( b ) * ( lp - 2112 + 21v - w l 2 )

5 6 ( b ) . (a + 2 * t) < 6(b) .

Hence Z [ m is Euclidean. (1, -1) is the set

of units of Z[-]. Suppose (20, yo) be an integer solution to the equation y2 + 2 = x3. In the ring Z [ G ] , we have (yo + Q)(yo - G) = zg. Since the integral divisor of yo + 2/--2 or yo - can only be f l , 20 is not a prime element in Z[.\/---z]. If a = m + n- is an irreducible divisor of 20,

ti = m - n&? is also an irreducible divisor of 20, and if alyo + n, then

b) By a), Z [ G ] is a unique factorization domain.

45

- alyo - loss of generality, yo +

from which it follows that a31yo + fl,c3ly0 - 1/--2. So without is of the form (m + nJ--2)3, rn, n E Z. Hence

yo = m3 - 6mn2 1 = 3m2n - 2n3 = (3m2 - 2n2) . n,

Thus it is easy to conclude that the integer solution to the equation y 2 + 2 = z3 are

y = -5 x = 3 and {

1302

Let p be a prime. Show that for any element a E Z/$, there exist b, c E Z/pZ such that a = b2 + c2.

Solution. (Harvard)

When p = 2 , it is trivial. Assume that p # 2. Let S be the set { b 2 I b E a/$}. Then S =

{o,T,22,. . . , (q)2} has exactly y elements. On one hand for any i # j ,

0 j), {o, T,z, elements in S. On the other hand, for any A2 E S where 9 < n < p , 0 < p - n 5 9 and A2 =

n2 = (n-p)2 = G . Thus S' = { b 2 I b EZ/$} = { a , T , ? 2 , . . - , ( q ) } has exactly 9 elements.

Now for any element a E Z/@, the set a - S := {a - b2 I b E Z/@} has exactly Since a - S C Z/pZ and Z/pZ has only p elements, (a - S ) n S # 0. Hence there exist 6 , c E a/@ such that a - b2 = c2 , that is a = b2 + c 2 .

-- 0 < j 5 q, 0 I: j 5 e, 2 i 2 - y 2 = i + j . i - j # O ( O < i + j 5 p - 1,

- , (q)2} are

2 2 -~

elements.

1303

For a ring R, R* denotes its multiplicative group of units, M,(R) the ring

(a) If a E R is nilpotent (am = 0 for some m) then 1 - a E R*. (b) Let J be a nilpotent ideal of R (J" = 0 for some m).

of n x n matrices over R, and GL,(R) = M,(R)*.

46

l a b

0 0 1 (d) Show that U = { ( 0 1 c )

Show that GL,(R) --t GL,%(R/J) is surjective, with kernel

a, b,c E F,} is a pSylow subgroup

- { A = ( a i j ) I 2 = diag{i,T,. . . , 1))

{ A E M,(R) I A = I mod M,(J)} =

= I + M , ( J ) .

(c) By calculating the possibility of the row vectors of A in GL3(Fq), it is easy to see that

U is a pSylow subgroup of GL3(3,). (e) We consider the field extension 3, C 3,s. There exsits a E 3,". Such

that Fi3 = (a) . Obviously, T, : F,s + 3,3, 3: H az is an invertible linear

47

transformation over Fq. Let A E GL3(Fq) be the matrix of T, relative to some base for Fq./Fq. Since .(a) = q3 - 1, the order of A in GL3(Fq) is q3 - 1.

( f ) Let R = Z/2M and J = U/2U < R. Then J 2 = 0 and % = R / J N

Z / U . By (b), the kernel of the surjective homomorphism

G&3(Z/2w) 3 GL3(Z/5Z)

is I + M 3 ( J ) . Obviously [ I + M 3 ( J ) J = 59

and IGL3(Z/U)I = 53(53 - 1)(52 - 1)(5 - 1) = 2 7 . 3 . s 3 . 31.

so (GL3(Z/2U)I = 2 7 . 3 . 512 - 31.

Let

P { A = ( a i j ) 3 x 3 E G&(Z/2U) 1 a ( A ) E U C GL3(Z/U)}

where cr is the homomorphism GL3(a/2u) -+ GL3(Z/U). Then P/ ker cr ‘v U and JP] = I kercr] . IUI = 512. Thus P is a 5-Sylow subgroup of GL3(Z/2U).

1304

Describe an infinite set of integral solutions (a ,b) of the Pell’s equation z2 - 2y2 = 1 (i.e., a , b E Z such that a2 - 2b2 = 1).

Solution. Suppose that (a, b) is an integral solution of x 2 - 2y2 = 1 . Then a must be

odd and b even. Let a = 2a’+ 1 and b = 2b’. Then we have bf2 = v. So must be a square number. Hence either both $ and a’ f 1 are square

numbers or a’ and If $ and a’ + 1 are square numbers, say a’ = 2b;, a’ + 1 = a; for some

integers ao, bo, then b” = b; . ad, (ao, bo) is a solution of z2 - 2y2 = 1 , and

&om the above consideration, it is easy to see that if (ao, bo) is a solution of z2 - 2y2 = 1, then (al = 2 4 - 1 , b = 2aobo) is another solution of z2 - 2y2 = 1 .

Obviously, ( 3 , 2 ) is an integral solution of x2 - 2y2 = 1. So if we take a. = 3 , bo = 2 and a, = 2aP-1 - 1 , bi = 2a,-lbi-l (i 2 l), we get an infinite set of integral solutions {(ai , bi) 1 i 2 0) of the pell’s equation x2 - 2y2 = 1.

(Indiana)

are square numbers.

a = 2 4 - 1 , b = f2aobo.

48

Remark. If a' and are square numbers, say a' = ug, a'+ 1 = 2bg, then (ao, bo) is an integer solution of z2 - 2y2 = -1 and a = 2 a i + 1, b = k2aobo. Conversely, if (ao,bo) is an solution of 2' - 2y2 = -1, then (a = 2ag + 1 , b = f 2 a o b o ) is an solution z2 - 2y2 = 1. For example, from the solution (7 ,5) of x 2 - 2y2 = -1, we get a solution (99,70) of z2 - 2y2 = 1.

1305

Let p be a prime. Let R be a commutative ring in which u p = a for all

(a) If R is an integral domain, prove that R is isomorphic (as a ring) to

(b) In the general case (i.e., R not necessarily an integral domain), prove that R is isomorphic (as a ring) to a subring of a direct product (not necessarily finite) of rings each of which is isomorphic to Z/pZ.

Solution. (a) Suppose that R is an integral domain. For any a (f 0) E R, since

UP - a u(aP-1- 1 ) = 0, ap-' = 1. Hence R is a field, and for any a E R, a is a root of the polynomial zP - z = 0. It follows that R has at most p elements. Since p is a prime, Char(R) = p and R = ZlpZ.

(b) Since for any a E R, u p = a, the nil radical of R is 0, that is, the intersection of all prime ideals of R is 0. Hence

a E R.

(Indiana)

all prime P

For any prime ideal P of R, RIP is an integral domain and for any a E RIP, - p - z - - a - to a subring of a direct product of rings, each of which is isomorphic to a/$.

- a. By (a), RIP -Z/pZ. Thus we have proved that R is isomorphic

1306

Let R be a commutative ring with 1. If R satisfies the a.c.c. (ascending chain condition) for finitely generated ideals then show that R satisfies the a.c.c. for all ideals. Give example (without proof) of such a ring which is an integral domain but not a p.i.d.

(Indiana)

49

Solution. If R satisfies the a.c.c. for finitely generated ideals, then for any ideal I of

R, 1 is finitely generated. For otherwise, we can construct a strictly ascending chain of finitely generated subideals of I . Hence R is a Noetherian ring, i.e., R satisfies the a.c.c. for all ideals.

X[z] , the polynomial ring over Z, satisfies the a.c.c. for all ideals, but not a p.i.d.

1307

Let R be a commutative ring. Let A be an ideal of R. (a) Show that S = (1 + a I a E A} is a multiplicatively closed subset of R. (b) Show there is a one-to-one correspondence between the prime ideals of

(c) If A is contained in every maximal ideal of R, what is S- lR? (d) If A is a maximal ideal of R, what can you say about the structure of

S - l R and those prime ideals P of R such that P + A # R.

S-lR? (Indiana)

Solution. (a) Obviously, 1 = 1 + 0 E S . For any 1 + a and 1 + b, a , b E A,

(1 + a ) ( l + b ) = 1 + (u + b + a b ) 6 S.

It follows that S is a multiplicatively closed subset of R. (b) It is well known that there is an one-to-one correspondence between

the prime ideas of S-lR and those prime ideals P of R such that P n S = 0. Obviously, P n S # 0 if and only if P + A # R here. This is what we need to prove (b).

J ( R ) , the Jacobson radical of R. Hence S-lR = R.

(d) If A is a maximal ideal of R, then S-lA is the unique maximal ideal of S- lR by (b). So S- lR is a local ring, and it is easy to see that S- lR is isomorphic to RA, the localization of R at the maximal ideal A.

(c) If A is contained in every maximal ideal of R, then A So every elements in S is invertible in R.

1308

Let I be a nilpotent ideal in a ring R, let M and N be R-modules, and let f : M -+ N be an R-homomorphism. Show that if the induced map

50 - f : M / I M ---t N I I N

is surjective, then f is surjective.

Solution. Since -

f : M / I M -+ N / I N

is surjective, f ( M ) + I N = N . It follows that

(Harvard)

I . N/f(M) = I N + f (M)/f(M) = N / f ( M ) .

N / f ( M ) = I . N / f ( M ) = 1 2 . N / f ( M ) = . . * = I n . N / f ( M ) = * . . = 0,

Hence

because I is nilpotent. So N = f (M) and f is surjective.

1309

Let F be a finite field containing 5 elements. Let t be transcendental over F . Explicitly construct one non-archimedean absolute value I I on F ( t ) . If is the completion of F ( t ) with respect to 1 1, show that the set of a(t) E F ( t ) satisfying la(t)l 5 1 is a local ring.


Let p = p ( t ) = t - 1 E F[ t ] . We define V, : F ( t ) -+ I?i by %(O) = 00 and Vp(f(t)) = Ic if 0 # f ( t ) = ~ ( t ) ~ * b ( t ) / c ( t ) , where b( t ) , c ( t ) E F[ t ] and

-

( b ( t ) , P ( t ) ) = 1 = ( .( t) ,P(t)) .

Obviously, we have i) V,(f(t)) = 00 if and only if f ( t ) = 0, ii) V,(f(t) . dt)) = V,(f(t)) + % ( S W and 4 v,(f(t) + d t ) ) L fin(Vp(f(t)>, VP(dt))) .

Then, by defining I f ( t ) l , = 2 - V p ( f ( t ) ) , we get an absolute value 1 1, (called padic absolute value) on F ( t ) . Obviously, I 1, is non-archimedean (If(t) +

Suppose F ( t ) is the completion with respect to 1 I p . Let R = {a(t) 1 a(t) E W l P I maXMt)l,, Is(t)lPH- - F ( t ) , Ia(t)lp 5 1) and M = {a ( t ) E R I la(t)l < 1). Since

14 + ml, L m={I4t>Ip, IP(t>lP)

51

and b ( t > P ( t ) l P = l+)lPIP(t)lP> -

R is a subring of the field F ( t ) and M is an ideal of R. - For any a(t) E R\M, we have la(t)lP = 1 and Ia(t)-'lP = 1 where a ( t ) - l E F ( t ) is the inverse of a@). So, a ( t ) - l E R and a(t) is invertible in R. Thus R is a local ring with maximal ideal M .

1310

Prove that the ideal generated by X I , . . . , X , in the polynomial ring a'[X1, . . . , X , ] cannot be generated by fewer than n elements.

Solution. Suppose that { Y l , Y2, * . . , Y,} is another generating subset of the ideal

( X I , X 2 , . . . , X , ) of the ring C [ X 1 , X2 , - . - , X,]. We have to show that m 2 n. We can write

( StanfonE)

x; = UilY1 + * * * + U l m Y m + fa

for any X ; where a i l , a j 2 , - . . , a i m E a', and f j is a sum of monomials in Y1,. . . , Y, of degree two or greater. In the same way, we also have

y j = b j iX i + bj2X2 + * * * + bjnX, + gj

for j = 1 ,2 , . . , rn, where b j l , . . . , bj, E a', and g j is a sum of monomials in X I , X z , .

so , X , of degree two or greater.

= 5 (5 u i j b j k ) xk + (terms in X I , . - . x , of degree 2 2) k=l j = 1

( i = 1 , 2 , . . . , n ) . Since X I , X2 , . . . , X , are algebraically independent over a',

m

a i j b j k = 6 i k (i, k = 1,2, * * * , n). j = 1

52

1311

Let A be a commutative ring, and let I and J be two (proper) ideals such that every prime ideal of A contains either I or J but no prime ideal contains both I and J . Prove that

A N A1 x A2

for some (nontrivial) rings A1 and A2.

Solution. Since every prime ideal of A contains either I or J but no prime ideal

contains both I and J , we have I J N ( A ) (the nil radical of A ) and I+J = A. There exist a E I , b E J such that a+ b = 1, and since ab E I J C N ( A ) , there exists an integer n such that (ab)" = an b" = 0. Let 1 1 = (a") and 12 = (b"). Then I1 and I2 are proper and I1 + I2 = A , since

( Stanford)

1 = ( a + b)2n E (a") + (b") = I1 + 12.

By Chinese Remainder Theorem, we have

I1 n 12 = 1 1 I2 = (a") . (b") = (a"b") = 0

and thus A N A/I1 x A/I2 where A/I1 and A/I2 nontrivial.

1312

Define f : [0, 1) -+ [0, 1) by

if 0 5 22 < 1, if 1 5 22 < 2. f ( x ) = { 2x,

22 - 1,

Find all z such that f(f(f(f(f(f(z)>>))> = 2.

(Indiana)

53

Solution.

We denote { u } = a - [u] here. Then it is clear that For any real number a , [u] denotes the largest integer 5 a (Gauss function).

f (z ) = 22 - [2z] = ( 2 2 )

f(f(f(f(f(f(4 * * .) = (2 * (2 . (2 . (2 * (2 . (22) . - 0 ) .

for any z 6 [0, 1). Hence

Since for any real number a and positive integer n, we have

{ n . { u } } = n { u } - [ n . { u } ] = n(u - [u]) - [n(u - [u])]

nu - [nu] = {nu} ,

f(f(f(f(f(f(.)))))) = { 2 6 4 = (642).

= nu - n . [a] - [nu - n[u]] = nu - n[u] - ([nu] - n[u] ) =

Notice that 2 = (642) = 642 - [64 . z] if and only if 63z = [64z]. Since z E [0, l), 632 = [64z] = [63z + z] (to be an integer) if and only if

1 2 63

Thus f(f(f(f(f(f(z)).--) = z if and only if z = &, 0 5 k 5 62.

1313

Let T be the ring of all real trigonometric polynomials

a, cos nz + b, sin nz. N

f( z) = a0 + n = l

Define deg f (x) = N where U N or bN # 0. Show that deg f . g = deg f + deg g. Use this result to prove that T is an integral domain which is not a unique factorization domain.


By using the orthogonality of the set ( l , cosnz , s innz I n E nV}, it is easy to see that

N

f ( z ) = a0 + a, cos nz + b, sin n2 0 n = l

if and only if all the coefficients of f(z), i.e., ao, a l , . . . , aN and bo, b l , - . , bN

are zero. Let

N

f (z ) = a0 + an cos nx + b, sin nz n= 1

and M

g(z) = c g + c cm cosmz + d , sinmz. m=l

Suppose that deg f(z) = N and deg g(z) = M . Since

(an cos n x + bn sin nz ) + (cm cos m x + dm sin mz)

+ c andm - 2 bncm ) sinkx] . m-n=k

The coefficients of cos(N+M)x and s in(N+M)x in f(z).g(z) are a N c M ~ b N d ~

and a N d M : b N C M respectively. If both of them are zero, then

a N . C M . d M = b N . d$ = - b N C M , 2

and so b N ( c $ + d&) = 0. Since c h + d& # 0 , bN = 0. Then we have a N C M = U N d M = 0. Hence a N = 0, which is contrary to aN or bN # 0.

Thus we have proved that

deg(f(z) .g(x)) = N + M = d e g f ( z ) + d e g g ( z ) .

Now f ( z ) . g(x) = 0 can happen only if either f(z) = 0 or g(z) = 0 by the

If f(z) . g(z) = 1, then the degree relation implies that degf (z) = 0 = above degree relation. So T is an integral domain.

degg(x). Hence the units of T are {*1}.

55

Obviously, in T we have

cos 22 = =

(cos z + sin z)(cos z - sin z)

(1 + a s i n z)(1- a s i n z).

Again by the degree relation, all the factors cosz f sin%, 1 f f i s i n z are irreducible, and cos z f sin z and 1 f d s i n z are not associates. Thus cos 2% in T does not have an essentially unique factorization into irreducible elements. Hence T is not a unique factorization domain.

1314

Let A be a Noetherian integral domain integrally closed in its field of fractions K . Let L be a finite separable field extension of K . If B is the integral closure of A in L, outline a proof that B is Noetherian.

Solution. Let p = Char(K)

and n = [L : K ] . If p = 0 or ( p , n ) = 1, it is easy to see that nLIK # 0 ( n ~ / K ( b ) = nb for any b E K ) . Now, suppose that p # 0 and p I n. Since L _> K is a finite separable extension, 1; = K[a] for some a E L. Denote a = a1,a2,- . . ,an be the set of the conjugates of a. For any f(a) E L,

n L / K ( f ( a ) ) = C f(ai). Let zn - clzn-l + c ~ z ~ - ~ - . * * + (-l), - c, be the

minimal polynomial of a over K and j be the minimal positive integer such that p ] j and cj # 0. Then by using Newton's identities on the elementary symmetric polynomials, we get

( Stanfod)

First, we claim that the trace function ~ L I K # 0.

n

i = l

TrL,&j) = ( - l )n+j+l * j - cj # 0.

Thus ~ L I K # 0. Let 211, u2, - . . , u, be a basis of L over K contained in B. The bilinear func-

tion (2, y) -+ T r L / ~ ( z y ) is non-degenerate since ' I ~ L I K # 0. Let ~ 1 , 2 1 2 , . . . , v, be the dual basis of ul , u 2 , - * - , u, (the elements in L satisfying T T L / ~ ( u ~ v ~ ) = 6 , j for all i, j). Then, for any z E B , 3: has the form lclvl + k2v2 + . . * + knvn ( k i E B) . Since xu, E B , Ici = ~ L / K ( ~ u , ) E A for any i. Hence B C Avl + Av2 + . + Av,. Since A is Noetherian, B is Noetherian.

56

1315

If A is a commutative ring and A[[%]] is the ring of formal power series over

A , show that if f = c anzn is nilpotent, then all the elements aa E A are

nilpotent. If A is Noetherian, prove the converse, i.e., that if all the elements

00

n=O

ai are nilpotent, then f is nilpotent.

Solution. 03

(Stanford)

Suppose that f = c anzn is nilpotent. It is easy to see that f" = 0

implies that a? = 0. So, first we get that a0 is nilpotent. Assume that n=O

ao, ai, . . . , ak is nilpotent. Since A is commutative, a0 + a12 + . . . + a k Z k is nilpotent and

is nilpotent. Hence 2 induction, all the elements ai are nilpotent.

Conversely, suppose that A is Noetherian and all the elements ai are nilpotent. Let I be the ideal generated by {ai I 0 5 i < m}. Since A is commutative Noetherian, I is finitely generated and nilpotent. If I" = 0, that is, b lbz - - . b, = 0 for any b l , b z , * , b, E I, then it is easy to see that f" = 0. Hence f is nilpotent.

is nilpotent and so ak+l is nilpotent. By n = k + l

1316

Let F be a field, and for n 2 1 let R, be the subring of the ring of polynomials F[x] consisting of polynomials

f0 + f i x + fizz + * . + fkxk E F [ z ]

such that f1 = fi = ... = fn = 0. (a) Show that R, is a Noetherian ring. (b) Show that the field of fractions of R, is F ( z ) , and determine the integral

closure of R, in its field of fractions. (Stanford)

57

Solution. (a) Since F [ x ] i s Noetherian and

is finitely generated as &-module, R, is a Noetherian ring by Artin-Tate Lemma (or Eakin's Theorem).

(b') Observe that

for any a ( x ) / b ( x ) E F ( x ) . It is clear that the field of fractions of R, is F ( z ) . Obviously, z is integral over &, since z is a root of Xn+' - zn+' E R n [ X ] .

Hence F [ x ] is integral over R,. On the other hand, if a(x ) /b (x ) E F ( z ) is integral over R,, a ( x ) / b ( z ) is integral over F [ z ] . Since F [ x ] is an integrally closed domian, a ( z ) / b ( z ) E F[z] . Hence F[x] is the integral closure of R, in its field of fractions.

1317

Describe all subrings of &. (A subring contains 1 by definition.) ( Stanford)

Solution. Let R be a subrings of&. Then R _>a by definition. Let

Obviously, S is a multiplicatively closed subset of Z containing 1. Let Zs be the localization of Z at the multiplicatively closed subset S . For any q / p E as ( q E Z, p E S),

On the other hand, for any q / p E R, we may assume that (p , q ) = 1 and pl + q k = 1 for some I , k E Z. Then

= q . E R. Hence Zs C R.

4 P P P

- 1 + k * - E R. 1 p l + q k -

So p E S and q / p E Zs. Hence R C Zs. It follows that R = Zs where

1 s = { O # P E Z ) - E R } . P

Thus (27s I S is a multiplicatively closed subset of Z}

is the set of all subrings of& (S can be choosed as the complement in Z of the union of some prime ideals of a) .

SECTION 4 FIELD AND GALOIS THEORY

1401

1) Let X be a finite set and G a subgroup of the group of permutations of X. Define a relation - on X by requiring x - y if either x = y or the transposition (2, y) (which interchanges z, y E X and leaves all other elements fixed) is an element of G . Show the following.

(a) - is an equivalence relation. (b) If G acts transitively, then all equivalence classes are distinct and con-

tain the same number of elements. ( c ) If Card(X) is a prime number and if G acts transitively and contains

a t least one transposition then G must be the whole permutation group of X. 2) Suppose f E &[z] is irreducible and has degree p , a prime number. If f

has exactly p - 2 real roots and 2 complex roots, show that the Galois group of f over & is the symmetric group S, on p symbols. Show that the polynomial

(x2 + 4) . x . (z2 - 4)(x2 - 16) - 2

is irreducible and determine its Galois group over &.

Solution. ( Cohmbia)

1) (a) By the defintion of -, - is reflexive and symmetric. Since

(x,Y)(Y,Z)(x,YY = ( z , z ) , (. # Y , Y # z) ,

it is easy to see that - is transitive. Hence - is an equivalence relation. (b) Let { z l , x 2 , ~ ~ ~ , z n } and { y 1 , y 2 , - - . , y m } be two equivalence classes of - determined by z = x1 and y = y1 respectively. Since G acts transitively on

X, there exists g E G such that g(x) = y. For any 1 < i 5 n, we have

(g(zl),u(zi)) = ~ ( 2 1 , ~ i ) g - ' E G .

Thus { ~ ( z l ) , 4z2), . e . , ~(2,)) are n distinct elements belonging to the equivalence class determined by y1 = g(x1). Hence m 2 n. Similarly, we have n 2 m. Thus m = n. So all equivalence classes contain the same number of elements.

60

(c) Suppose G acts transitively and contains a t least one transposition. Then, by (b), all the equivalence classes contain the same number of elements, say n, and n > 1. So Card(X) = m . n, where m is the number of distinct equivalence classes determined by -. Since Card(X) is a prime, m = 1 and n = Card(X). Thus for any z,y E X , z - y, i.e., (z,y) E G. It follows that G is the whole permutation group of X , which is generated by all the transpositions.

2) Suppose P

f(.) = rI(z - Ti) i = l

in @[XI. So E = & ( T I , . . . , rP) is a splitting field of f(z) over & contained in a. Identify G = Gal(E/Q) with a permutation group of the set X = { T I , .. , r p } of the (distinct) roots by q + q1x. For any ri,rj E X, since f(z) is irreducible and f ( r i ) = 0 = f ( r j ) , there exists an isomorphism of&(r,)/& into&(rj)/& by sending ri to r j . Since E =&(rl, , r p ) is a splitting field of f(z) over &(ri) , and also over Q ( r j ) , this isomorphism can be extended to an automorphism q of El&. Then q E Gal(E/&) and q(ri) = ~ j , which shows G acts transitively on X .

Consider the conjugation automorphism onG. This maps f(z) to itself. Let r1 and r2 be the two non-real roots of f(z). Thus the conjugation interchanges r1 and r2 = Fl and fixed all other real roots. Hence the restriction of the conjugation to E is an element of G and it is a transposition. Thus the Galois group G of f (z ) over & is the symmetric group Sp on { T I , .. + , r p } .

By Eisenstein criterion,

f(z) = (z2 + 4)2(z2 - 4)(x2 - 16) - 2

is irreducible over &. Let

g ( 2 ) = (z2 + 4)2(z2 - 4)(z2 - 16).

The real roots of g(z) are 0, f2, f4, and the graph of y = g(z) has the form

Fig.l.1

61

Since I g ( n ) l > 2 for any odd integer n, it is easy to see that f(z) = g(z) - 2 has five real roots and two non-real roots. Hence the Galois group of f(z) over & is ST.

1402

Let p be an odd prime. Let C p be a primitive pth root of unity and g a primitive root ( mod p ) (i.e., g is a generator for (Z/pZ)*). Fix e, a divisor of p - 1 and put f = ( p - l) /e. Define

f -1

.qa = C(<p)gej+i j =O

Show that, for any i, vi generates a subfield of & ( C p ) of degree e over &. Hint. Use the generator CT : C p --+ ( C p ) g of the Galois group of&(Cp) over &.

(Columbia)

Since g is a primitive root ( mod p ) , CT : C p H (&,)g is an automorphism of Solution.

&(GI over & and

G = Gal(&(<,)/&) =< CT >= {c, CT’,. , up-’, aP-’ = 1 1. Obviously, {a(<p),a2(~p),~~~,~P-1(<p) = C p } is a base for&(<p)/&.

Let H =< C T ~ >< G. Then JHJ = f,

@ ( c P ) : InvH] = IHI = f

and [Inv(H) : &] = e . Since H is normal in G, Inv(H)/& is a Galois extension and Gal(Inv(H)/&) 21 G/H.

Now for any i,

62

and

f --I

j = O

f - I

j=1

f - 1

. . .

e- 1 are distinct. Hence n (z - q j ) is the minimal polynomial for qi over Q (for any

j = O

i). It follows that 71; is a primitive element for Inv(H)/Q, i.e., Inv(H) =@(qi) for any i. Thus qi generates a subfield of &(&) of degree e .

1403

Let K be a field and z be an element of an extension of K such that 2 is transcendental over K . Put G = Aut(K(z)/K), and let H denote the subgroup of G consisting of the substitutions 2 + 2 + b with b E K .

(a) Let A , B E K [ X ] , AB 6 K , gcd(A,B) = 1, and put y = A(z)/B(z) . Show in succession that the polynomial A - yB E K ( y ) [ X ] is not 0, that z is

63

algebraic over K ( y ) , that y is transendental over K , that A - y B is irreducible in K ( y ) [ X ] , and that

[ K ( z ) : K ( y ) ] = max{deg A, deg B}.

(b) Show that the elements of G are given by the fractional linear substi-

(c) Show that when K is infinite, then G is infinite and the fixed field of G

(d) Show that when K = lFq and put

tutions z -+ a z + b/cz + d with a , b, c , d E K and ad - bc # 0.

is K . Find the fixed field of H .

z (zq2 - z)q+l/(z' - z)qZ+l,

then ordG = q3 - q and the fixed field of G is n;h(z). Conclude that n;h(z) is a Galois extension of a given field L with lFq C L C Fq(x) if and only if L 2 lFq(z ) . Find the fixed field of H .

Solution.

in K [ X ] . Suppose A - yB = 0 in K ( y ) [ X ] . Then

( CoZumbia)

(a) Since gcd(A, B ) = 1, there exist S, T E K [ X ] such that A S + B T = 1

It follows that degA = 0 and deg B = deg(A) = 0, which is contrary to

Since A(z) - y B ( z ) = 0, z is algebraic over K ( y ) . If y is algebraic over K , z must be algebraic over K , which is contrary to the assumption. Hence y is transendental over K .

Again, since gcd(A, B ) = 1 in K [ X ] (c K ( y ) [ X ] ) , A - y B is irreducible in K [ y ] [ X ] = K [ X ] [ y ] . Thus A - yB is irreducible in K ( y ) [ X ] . Thus the minimal polynomial of z over K ( y ) is the monic polynomial which is a multiple in K ( y ) of A - yB and

A B 6 K . Thus A - yB # 0.

[ K ( z ) : K ( y ) ] = deg(A - y B ) = max{degA,deg B} .

(b) By (a), y = A ( z ) / B ( z ) is a generator of K ( z ) / K (i.e., K ( y ) = K ( z ) ) if and only if max{ deg A, deg B } = 1, or if and only if y has the form az+ b/cz +d, ad - bc # 0.

(c) By (b), it is easy to see that G N G L z ( K ) / K * .12, where

K* . I2 = {diag(a,a) 1 a E K*}.

64

If K is infinite, so is G. Obviously, K C Inv(G), the fixed subfield of G. On the other hand, if

there exists some y = f(z)/g(z) E Inv(G)\K, then by (a), K ( z ) / K ( y ) is a finite dimensional simple extension. Hence AutK(y)K(z) is finite, which is contrary to the facts that G c AutK(y)K(z) and G is infinite. Thus we have Inv(G) = K .

Similarly, we have Inv(H) = K , since H is also infinite. (d) Suppose K = Fq. Then

and [F,(z) : Inv(G)] = q3 - q .

For any g : 2 H az + b/cz + d, (ad - bc # 0)

in G, it is routine to check

g(z) = g((zq2 - Z)Q+1/(2~ - 2)q2+1) = 2.

Hence F (z ) c Inv(G). On the other hand, we have

(XP2 - x)q+l/(XP - X)q2+1

1 (xq2 - X) = ( (xq-x) )q+l . (XP - X)+P

(1 + xn-1 + x ( n - 1 1 2 + . . . + X(P-Q?)q+l

(XP - X)n”-n - -

Denote A = (1 + XP-’ + X(P-1)2 + . . . + X(Q-1)Q)q+l

and B = (Xq - X)q2-g.

Then gcd(A, B ) = 1 and

max{deg A, deg B } = q3 - q .

Hence we have [Fq(z) : Fq(z)] = q3 - 4. Thus Inv(G) = Fq(z) and Fq(z) is a Galois extension of a given field L with FP C L C F,(z) if and only if L _> F(z) .

65

Similarly, we have [Fq(z) : Inv(H)] = q since IHI = q. Let A = Xq - X and B = 1 in F,[X] and let z' = zq - z E Fq(z). Then z' E Inv(H) and by (a), [Fq(z) : Fq(z')] = q. Thus

Inv(H) = F,(z') = Fq(zq - z).

1404

Let E = C(Y) with Y an indeterminate, F = C(2) with Z = Y" + Y-",

(a) Show that there are unique automorphisms u and r of El@ such that a ( Y ) = CY and T(Y) = Y-l , and that the subgroup G of Aut(E) which these generate is isomorphic to D,, the Dihedral group of order 2n.

(b) Show that Y is a root of a polynomial of degree 2n with coefficients in F.

(c) Show that EIF is a Galois extension with Galois group G.

and < = earriIn.


(a) Since Y is a generator of E over C, there are unique homomorphisms u and T of E to E over C such that ,(Y) = CY and T(Y) = Y - l . Obviously, (T" = 1 = r2. Hence c and T are automorphisms of E/C. Since ord(u) = n, ord(7) = 2 and Tu = u"-'T in G, G =rx D,, the Dihedral group of order 2n.

(b) Since

X2" - 2.x + 1 = X2" - (Y"+Y-") + 1 = (X" - Y")(X" - Y-") E F[X],

Y is a root of X2" - ZX + I. (c) Obviously, E/Inv(G) is a Galois extension with Galois group G and

[E : Inv(G)] = 2n. Since u(Z) = u(Yn + Y-") = Z and T(Z) = 2, we have Z E Inv(G) and F = C(Z) Inv(G). I t is readily verified that X2" -ZX+ 1 is irreducible over F (Z is transendental over C and X2" - Z X + 1 is irreducible in C[Z][X],for example, using 1403). E is a splitting field of X2" - ZX + 1 since

x 2 n - zx + 1 = ( X - Y)(X - C Y ) - . . ( X - C"-'y) (X - y-')(x - (Y-') * ( X - y-1y-l)

66

in E . F = Inv(G) and E / F is Galois with Galois group G.

It follows that E / F is Galois and [E : F ] = 2n. Hence we have

1405

Let F = &(z) be the field of rational polynomials in one variable x over & (i.e., the quotient field of the polynomial ring&[z]). Consider the elements (T, 7 in A u b ( F ) (i.e., field automorphisms of F ) given by ( ~ ( z ) = 2 - z and 7(z) = A.

2- 1 (a) Find the subgroup G (of Aub(F)) generated by (T and 7. (b) If K is fixed field of G in F , find a finite subset S of F such that

(c) How many subfields lie strictly between K and F ? How many of these K = &(S).

are Galois over K? Justify your answers.

Solution.

r2 = 1 and UT = TU. Hence

(Indiana)

(a) Since (~’(z) = z, ?(z) = z and T(T(Z ) = = (TT(z), we have (T’ = 1,

where K 4 is the Klein 4-group. (b) Obviously U ( Z - 1) = 1 - x and ~ ( z - 1) = 5. It is easy to see that

(z - 1)2 + (x11)2 E K , the fixed field of G in F . Denote 7 = (z - 1)’ + &. Then&(q) 2 K F . Let

f ( t ) = t4 - qt2 + 1 € & ( V ) [ t ] .

Then f ( t ) is irreducible in &(q)[t] and F is a splitting field of f ( t ) since

in F [ t ] . Hence&(q) C F is a Galois extension and [ F : &(71)] = 4. On the other hand, K C_ F is a Galois extension and [F : K ] = IGI = 4. It follows that

(c) By the fundamental theorem of Galois theory, there are exactly 3 subfields lying strictly between K and F which correspond to the three proper

67

subgroups of G K4. Since all the subgroups of K4 are normal, all these 3 subfields are normal over K (finite dimensional and separable). Thus all these 3 subfields are Galois over K.

1406

Consider &(t), the field of quotients of the polynomial ring @[t]. Let u and T be elements of Autq(&(t)) given by u(t) = 3 and ~ ( t ) = -t. Let G be the subgroup of Aub(&(t)) generated by u and T.

(a) Identify G. (b) Let H be the subgroup of G generated by u2 and T . Find a E @(t) such

that &(a) is the fixed field of H (Justify your answer).

Solution.

t . It is easy to see that r2 = 1, u4 = 1 and T(T = C3T. It follows that G is isomorphic to the Dihedral group D4.

(Indiana)

(a) Obviously, T 2 ( t ) = t , u2(t) = u (E) = - f , CT 3 ( t ) = -3 and a4(t) =

(b) Since

the fixed field of H . Hence,

and obviously [&(t) : InvH] = (HI = 4.

On the other hand,

1 = (2 - t 2 ) (2 - F )

= (z - t ) ( z + t ) (x - ;) (. + :) is irreducible over & (t2 + $) and Q ( t ) is a splitting field over & ( t 2 + $) of separable polynomial f(z). So

68

is a Galois extension and

It follows that

that is,

[&(t) :& ( t 2 + $)I = 4.

InvH =& (t2 + $)

1407

Let & denote the field of rational numbers. let K =&(a) , where a is a root

a) Prove that f(x) is irreducible over &. b) Prove that K/& is Galois. Hint. Consider a' - 2. c) Find a generator of the Galois group Gal(K/&).

off(.) = x3 - 3x + 1.

(Indiana) Solution.

a) If f(z) = x3 - 32 + 1 is reducible over &, then f(z) has a factor with degree one in &[.I, that is, f(x) has a root in &. Since f(2) = x3 - 3a: + 1 is monic, the rational roots must be integral factors of 1. But fl are not roots of f(z). This is a contradiction. Thus f(x) is irreducible over &.

b) Let ,B,r be the other two roots of f(2) = z3 - 32 + 1 in a splitting field of f(x) over K. Obviously, cr + p + 7 = 0, crPy = -1 and the discriminant A of f(2) is 34. Hence /3 + y = -a and

69

It follows that (Y, = -a2 - (Y + 2 and -A - (Y = a2 - 2 are the roots of f(x). So K is a splitting field of f ( x ) . Hence K / Q is Galois since Char(&) = 0. (We may check a2 - 2 is a root of f ( x ) directly by using the hint).

c) From a) and b), it is easy to see that the Galois group Gal(K/&) F A3.

( T : K = & ( a ) + K a k + ( Y 2 - 2

is a generator of Gal(K/&).

1408

Let K be a field and 3: an indeterminate. (a) Show that the rational functions f a =

(b) As K-modules, K [ z ] and K ( z ) have what dimensions? (c) Let G denote the additive group of K, acting on K ( x ) by a E G sending

z to z + a. Assume that K is infinite. Let f E K ( z ) . Show that the G-orbit of f spans a finite dimensional K-module if and only if f E K[z].


1 a E K } is linearly dependent over K. There exist some non-zero elements (Y~,-..,(Y, E K and some distinct elements a l , u2,. . . , a, E K such that

(a E K ) are linearly independent over K.

(a) Suppose that {fa =

n

Hence

and

i# i

which is not true. Thus { f a = -& I a 6 K } is linearly independent. (b) dimK K[x] = dimK K ( z ) = 00.

(c> Suppose

f (x) = a,zn + an-lxn-l + ' - * + a12 + a0 E K [ z ] .

70

We claim that for any n + 1 distinct elements al,a2, ". ,an+l in K (note that K is infinite), f ( x + a l ) , f ( z + a2), . . . , f ( x + an+l) spans the subspace generated by the G-orbit of f .

For any a E G, we take (,01,/32,.'-,/3n+l) E K n t l to be the solution of the equation system

1 1 - 0 . 1 a1 a2 . * f a n t 1

a: a; ... a;+1 . . . . . . . . . . . . i a; a; ... a;+1

Then it is clear that

= P ~ ( Z + $ I ) ~ + P z ( x + a 2 ) " + . * . + P n + l ( ~ + a n + l ) ~

and also

( Z + .)a = P ~ ( Z + alli + /32(z + a2)a + . .. + Pn+l(z + Qn+1li

for 1 5 i < n. Hence

f(.+a) = P l f ( z + a l ) + P a f ( z + a 2 ) + . . . + P n + l f ( 2 + ~ 1 n + l ) .

This shows that { f(x + ai) 1 1 5 i 5 n + 1) spans the subspace generated by the G-orbit of f .

On the other hand, suppose f ( z ) E K ( z ) , and the G-orbit o f f , { f (z+a) 1 a E K } spans a finite dimensional K-module. We write f(x) = g ( z ) / h ( x ) where g(z) , h ( z ) E K [ z ] and ( g ( z ) , h ( z ) ) = 1. Let f ( x + a l ) , f ( z+a2) , .. . , f(x +an) (ai E K ) span the subspace < {f(x+u) I u E K } >. Then for any a E K , there exist PI , P 2 , . . . , Pn in K such that

9(" + a ) h(x + u)

f ( x + a ) = n

g ( x + a i )

= - y P i h(x + ai) i=l n

71

Since (g(z -+ a) , h(z + u)) = 1, we have n

h(z +.)I n h(z +(.a)

i = l

for any a E F . Since K is infinite, an easy discussion in a splitting field of h ( z ) over F will lead to degh(z) = 0. Thus we have

1409

Let E be a finite Galois extension of F and let f(z) be an irreducible polynomial in F [ z ] . Show that all the irreducible factors of f (z) over E are of the same degree.

Solution. For any (T E G = Gal(E/F) , we still denote u to be the isomorphism

E[z] ---t E[z] which extends (T on E and maps x to z. Since f(z) E F [ z ] , a(f(z)) = f(z). Let e(z) be a monic irreducible factor of f(z) over E. Then, for any u E G, u(e(z)) is an irreducible factor of f (z ) over E.

We prove in the following that all monic irreducible factors of f ( z ) over E arise in this way. Thus all the irreducible factors of f (z ) over E are of the same degree.

Suppose e’(z) is another monk irreducible factor over E. Let a and a’ be roots of e(z) and e‘(z) in some extension field of E. Then, CY and a’ are roots of f(z), which is irreducible over F . Hence we have an isomorphism 7 : F ( a ) -+ F(a’) which sends a to a (u E F ) and CY to a’.

Since E / F is Galois, we can write E = F ( P ) , where ,B is a root of g(z), a separable irreducible polynomial over F . Obviously E is a splitting field of g(z). Then E(a) = F ( a ) ( P ) and E(a’) = F(a’) (P) are splitting fields of g(z) over F ( a ) and F(a’) respectively. Hence 7 : F ( a ) -+ F(a’) can be extended to an isomorphism 7 of E(a) onto E(a’). Note that y (P) may not be P, but q(P) is a root of q(q(z) ) = g(z). It follows that

Now, since a and a’ are roots of the monic irreducible polynomial e(z) and e’(z) over E respectively, 7 : E ( a ) + E(a’) is an isomorphism and ~ ( a ) = a’, we must have (vlE)(e(z)) = e’(z) and dege(z) = dege’(z).

( c o zum bia)

: E -+ E is in G.

72

1410

(a) Let K be a field of characteristic p > 0. Show that the polynomial tP - t - c in K[t] is either irreducible or splits completely into p linear factors over K.

Hint. If u is a root of t P - t - c then so is u+ 1. (b) Let F be the splitting field of the polynomial t62 - 1 over Zg. Show

Hint. First prove that the zeroes of t62 - 1 form a cyclic group G of order that [ F : & I = 3.

62.

S o h tion. Suppose that t p - t -c = f ( t ) . g ( t ) in K[t] where f ( t ) is a monic polynomial

of degree n, 1 5 n 5 p - 1. Let E be a splitting field of tP - t - c and let u E E be a root of this polynomial. Then for any m EZ,, the prime field of K ,

(Indiana)

f ( t ) = (t - u - il)(t - u - i2) * * * (t - u - in).

Comparing the coefficients of the term of degree n - 1, we obtain n . u + il + i2 +. . . + in E K . So we have n . u E K . Since p . u = 0 and there exist integers v and w such that w . n + wp = 1, u = (v . n + wp) u = v(n . u) E K . Thus we have

t p - t - c = ( t - u - m )

in K[t ] .

F" and G is cyclic of order 6 2 since (b) Let G be all the zeroes of t62 - 1 in F . Obviously, G is a subgroup of

(t6' - 1)' = 62t6' = 2t61 # 0.

It follows that [ F :as] 2 3.

73

Let E be a extension field of Z5 such that [E : Z5] = 3. Then E* is a cyclic group of order 53 - 1 = 124. Let GI be its unique subgroup of order 62. Then all the elements of G' satisfies t6' - 1. So t6' - 1 splits in E and E is a splitting field of t62 - 1. Thus we have E N F and [F :as] = 3.

1411

(a) Suppose you are given a field L , & C_ L a', such that L/& is algebraic and every finite field extension K / L , K 5 a' is of even degree. Show that every finite field extension of L must in fact have degree equal to a power of 2.

(b) Show that such a field L actually exists. (Indiana)

Solution. (a) Let K / L ( K C a ' ) be a finite field extension. We have to show [K : L]

is a power of 2. For this purpose, we may assume that K is Galois over L. Let G = GalK/L and JGJ = 2n . m where m is odd. By Sylow's Theorem, G has a subgroup H of order 2n. If K' is the corresponding subfield of K I L , then [K : K'] = 2" and [K' : L] = m. Since L has no proper odd dimensional extension field, we must have m = 1, and so K' = 1; and [K : L] = 2n.

(b) Let L be the field of real algebraic numbers, that is, the subfield of 1R of numbers which are algebraic over &. Then & 2 L & (X and L/& is algebraic.

Now for any finite field extension K I L , K a', there exists some element a: E K such that K = L ( o ) by Primitive Element Theorem. Let f(a:) be the minimal polynomial of o over L. Then f(a:) has no real root since f(a:) is irreducible in L[z] . So, f(z), when decomposed in nZ[a:], is a product of irreducible polynomials of degree 2. Hence,

[K : L] = [ L ( a ) : L] = degf(s )

is even. By (a), it is in fact a power of 2.

1412

Let K be a field of characteristic p # 0. The set {a:" J a: E K } is a subfield

(a) Let L be an intermediate field between Kp and K . If [L : KP] is finite, of K that is denoted by KP (no proof required).

prove that it is a power of p .

74

(b) A subset B of K is called pindependent if for any finite set bl, b 2 , . . . , b, of distinct elements of B

Prove that if KP # K, then K contains a maximal pindependent subset B. (c) Prove that the set B of part (b) satisfies Kp(B) = K .

(Indiana) Solution.

such that (a) If [L : KP] is finite, there exists a finite set of elements {bl , b2,...,bn}

KP(b1,b2,.*.,bn) = L. Without loss of generality, we can assume that bi $! KP(bl-..bi-1) for any 1 5 i 5 n (KP(b1,. + . , bi-1) = KP when i = 1). Since

and bi @ KP(bl,.. . ,bi-l),

[KP(bl,. . . , bz) : KP(b1, * . . , bz-l)] = p.

tP - bp is irreducible in KP(b1,. . . , bi-l)[t]. Hence

It follows that

[L : KP] 1 [KP(bl, b2, * * . , b,) : L] n

= n [ K P ( b l , - - - , b i ) : KP(bl,.. . ,b;-l)] i = l

= p”.

(b) If KP # K , there exist pindependent subsets. For example, if 6 E K\Kp, B = { b } is a pindependent subset of K . Now suppose that {Bi I i E I} is a chain of pindependent subsets of K . Let B = U Bi. Then for any

finite set { b l , . . . , b,} of distinct elements of B , there exists some i, such that { b l , b 2 , - - - , b m } C Bi. Since Bi is apindependent subset,

i € I

[KP(bl ,b2, . . - ,bm): KP) =pm.

It follows that B is pindependent. By Zorn’s Lemma, K contains a maximal pindependent subset.

75

(c) Let B be a maximal pindependent subset of K . Suppose Kp(B) c K . Let b E K\Kp(B). Then B U { b } is pindependent. The reason is that, for any distinct elements b l , b 2 , . - * , b, in B U { b } , if b @ { b l , b 2 , . . . , b m } , then

and [KP(bl, * * 1 , bm) : KP] = pm,

and if b E {bl,bz,...,b,}, say, b = b,, then {bl,-..,b,-l} C B and we still have

[KP(bl,. . . , b,) : KP]

[KP(bl, * . . , b,) : KP(b1, . . . , bm-l)] . [KP(bl,. . . , b,-l) : KP] = - - p.*"-l = p"*

Since B C B U { b } , the independency of B U { b } contradicts the maximality of B. Thus we have K P ( B ) = K .

1413

Let K be a finite field with pr elements ( p a prime) and n be a positive integer. If m is an integer which divides n and f(t) E K[t] is an irreducible polynomial of degree m, show that f divides t P r n - t .

Solution. Let E be a splitting field of t p r n - t over K . Then (El = prn and [E : K ] = n.

Let (Y be a root of the irreducible polynomial f ( t ) in some extension field of K. Then IK(a)l = prm and [ K ( a ) : K] = m.

E and L N K(a) . Hence there exists an element p E L C_ E such that f(P) = 0, that is, f ( t ) is the minimal polynomial of p. Since pp'" - p = 0, we have f ( t ) I ( t P r n - t ) .

( Indiana)

Since m I n, E contains a subfield L such that K & L

1414

Let K I F be a finite extension of fields and let L and E be intermediate fields, with E / F Galois and [K : L] = p , a prime. Prove that if p does not divide [E : F ] then E L.

( Indiana)

76

Solution.

splitting field. Let Let f(z) be a separable irreducible polynomial over F such that E is its

E * L = E(L) L(E) be the composite of E and L in K . Then E L = L(E) is a splitting field of f (z ) over L. Hence E + L is Galois over L. Let a E E be a root of f(z). Then E = F ( a ) and E . L = L(E) = L(a). For any

c E Gal(E. L/L) = Gal(L(a)/L),

it is clear that g ( ~ E Aut(E/L n E ) . And further, we have

Gal(E. L/L) N_ Gal(E/L n E ) .

Now suppose E prime. It follows that

L. Then E . L = K , since L C E . L and [K : L] = p , a

p = IGal(E. L/L)I = IGal(E/L n E)I

dividers IGal(E/F)I = [E : F ] , contrary to the assumption. Thus we have E C L.

1415

Let Ki be the subfields of a' defined as follows: KO = &. If i 2 0, Ki+l is the smallest subfield of @ containing the set

(0 €6' 10" E Ki for some n > 0).

Let

i=O

(1) Prove K is a field. (2) Let f(z) E K [ z ] be irreducible. Prove that deg(f) 2 5.

(Indiana) Solution.

(1) Since KO C_ K1 C_ C_ Ki C_ Ki+l C_ * * a

M

is a chain of subfields of a'. It is clear that u Ki is a subfield of a'. i= l

77

(2) (Remark: degf (z) may be 1). Let f(x) E K [ z ] be an irreducible polynomial with degf (z) > 1. There

exsits some i such that f(x) E Ki[x]. Suppose degf (z) 5 4. By the formulas for the roots of quadratic, cubic and quartic equations and

Ki+l 1 (0 E(C 10" E K , for some n > O } ,

f(z) splits in Ki+3[z], hence in K [ z ] . Contradicts the irreducibility of f(z). Hence deg f (z) 2 5.

1416

Let K be an extension field of Fp, the field with p elements. Let a be an algebraic element in K . Prove that [Fp(a) : Fp] is the smaIlest positive integer m such that

Solution.

E Fp, where g ( n ) = p"-l. P - 1

( I n d i a n a )

Let n = [Fp(a) : Fp]. Then IFp(a)I = p" and upn-' = 1. Since ( ~ ( g ( ~ ) ) P - l =

On the other hand, if apn-l = 1, a d n ) E Fp.

E Fp, for some positive integer m, then

a P m - l - - ( & 4 ) P - l = 1, I

m so u is a root of zp a E E. Then [E : Fp] = m and Fp

- x. Let E be a splitting field of zPm - z over Fp and Fp(a) C E. Hence

n = [F,(u) : F,]([E : pP] = m.

Thus [Fp(a) : F] is the smallest positive integer m such that E Fp.

1417

Let F 2 K be a field extension of finite degree m. Let f E K[t ] be an irreducible polynomial of degree n. If m and n are coprime then show that f remains irreducible in F [t].

Solution. Suppose that f ( t ) is reducible in F[t] and let f( t) = g( t ) -h( t ) in F[t] where

g ( t ) is a irreducible polynomial in F[t] of degree k, 1 5 k < n. Let E = F(cr),

( I n d i a n a )

78

where cr is a root of g ( t ) in some extension field of F . Then [E : F ] = Ic since g ( t ) is irreducible in F [ t ] . So

[E : K ] = [E : F ] [ F : K ] = Ic . m.

On the other hand,

[E : K ] = [E : K(cr)]. [K(a!) : K] = [E : K(cr)]. n,

since cr is a root of f ( t ) and f ( t ) is irreducible in K[t] . It follows that n I Ic -m, which contradicts (m, n) = 1 and 1 5 Ic < n. Thus f ( t ) is irreducible in F[t] .

1418

Find a Galois extension E over & with Gal(E/&) cyclic of order 16. (Stanford)

Solution. For any positive integer n, the cyclotomic field & ( z n ) over & is a Galois

extension and [&(zn) : &] = d(n) where z , is an n-th primitive root of the unit, d(n) is the Euler &function. It is easy to see that IGal(&(zi)/&)I = +(n) and Gal(&(z,)/&) N Aut(G) where G is the cyclic group of order n. When n is prime, Aut(G) is cyclic of order n - 1.

So if we take n = 17, E = &(z17), then E is a Galois extension of & with Gal(E/&) cyclic of order 16.

1419

Find a Galois extension E over & with Gal(E/&) cyclic of order 32. (Stanford)

Solution. As in 1418, for any positive integer n, the cyclotomic field &(zn) over Q is

a Galois extension and [&(zn) : Q] = +(n) where z, is an n-th primitive root of the unit, 4(n) is the Euler $-fur Ition. I t is easy to see that IGal(&(td)/&)I = 4(n) and Gal(&(z,)/&) E Aut( ;) where G is the cyclic group of order n. When n = 2" and m 2 3, it is well known that Aut(G) -.Zz @ . Z p - z .

By the Fundamental Theorem of Galois Theory, if we take E = Inv(&), then & C E is a Galois extension and Gal(E/&) 1: .Z2,-, .

Taking m = 7, then & E is a cyclic extension of order 32.

79

1420

Let E / F be a finite Galois extension, G = Gal(E/F) and a E E . Consider the F-linear map Ma : E + E, M,(z) = az. Show that its trace is given by Trp(M,) = C g ( a ) where


varies over G.

Let z be a primitive element of E / F , [E : F ] = n and

Then

is the minimal polynomial of z over F, and u1(z ) , uz(.z), .. . , gn((z) are distinct. Now, for any a E E , a has the form

since { l , ~ , . + , z n - l } is a base for E / F . To prove that

it suffices to prove that

for any 1 5 k 5 n - 1. Obviously, f(z) is also the minimal polynomial of the F-linear map M, :

E + E , M , (z) = z . 3c. Since f(z) has distinct roots C T ~ ( Z ) , gZ(z), - . . ,a,@) in E, the matrix of M , (E M n ( F ) ) , say, relative to the base (1, z , . - - , zn--l}, is similar to diag{ul(z),...,an(z)} in Mn(E). Anyway, we have

n

i = l

and for any 1 5 i 5 n - 1,

n

This completes the proof.

Part I1

Topology

a3

SECTION 1 POINT SET TOPOLOGY

2101

Let A and B be connected subspaces of a topological space X, such that A n B # 8. Prove that A U B is connected. If A and B are path connected, need A U B be path connected?

Solution. Let f be any continuous map from A U B to So = {-1,l). Since A is

connected, f ( ~ must be constant. Without loss of generality, we may assume that A c f-'(-l). By the same reason, f l ~ is also constant. Let zo E AnB. We have f(zo) = -1. Since f is continuous, there exists a neighborhood of zo, say U , such that U c f-'(-l). But since zo E B, there is a point of B which belongs to U . Therefore we have B c f-'(-l). Hence f is not surjective. It means that A U B is connected.

The following example shows that if A and B are path connected then A u B needs not to be path connected. Let

(Indiana)

A = ((0,O)) C R2

1 and

B = {(z,sin -) I 0 < z 5 1).

Then A and B are path connected and A n B = A , but A U B is not path connected.

3:

2102

Suppose that A and B are compact subspaces of spaces X and Y respectively, and that N is an open neighborhood of

A x B c X x Y .

Prove that there are open sets U c X and V c Y such that

84

A x B c U x V c N .

(Indiana) Solution.

Let' z be a point of A . For any y E B, since (z, y) belongs to A x B and N is an open neighborhood of A x B c X x Y , there exist open sets U,(z ) c X and V,(z) c Y such that

( Z C , Y ) E V,(z) x V,(4 c N .

Therefore the family of open sets {Vy(z), y E B } covers B. Since B is compact,

there is asubcover {VYi(z),i = l , . . . , p} such that B c U V,,(z). Let U ( z ) = P

i = l I) n fi Uyi(z) and V(z) = 0 V,;(z). It is obvious that U ( z ) and V(z) are open

sets of X and Y respectively and that i = l i = l

{z} x B c U ( z ) x V(z) c N .

On the other hand, { U ( z ) , z E A } is an open cover of A. Since A is also U

compact, there exists asubcover { U ( z i ) , j = l , . . - , q } such that A c u U ( z j ) .

Let U = U U ( z j ) and V = n V(zj). It is easy to see that U and V are open

sets of X and Y respectively and that A x B c U x V c N .

j = 1 q q

j=1 j = 1

2103

Let X be a locally compact Hausdorff space. Let A and B be disjoint subsets of X , with A compact and B closed. Does there exist a continuous function f : X -+ [0,1] such that f l ~ = 0 and f l ~ = l?

Solution. If X is compact, then X is normal and the existence o f f is obvious. Hence

we may assume that X is noncompact. We denote by X" the one-point compactification of X. Since X is locally compact and Hausdorff, X' is compact and Hausdorff, and consequently is also a normal space. Let X* = X U {m}. It is easy to see that A and F = B u{m} are two disjoint closed subsets of X'. Then by the Urysohn Lemma there exists a continuous function F: X' -+ [0,1]

( Cincinnati)

85

- - such that f l ~ = 0 and the requirements that f l ~ = 0 and f l ~ = 1.

= 1. Therefore, the restriction of f on X , f, satisfies

2104

No proofs, only the correct answers to the question asked, are required for

If X and Y are topological spaces, the join of X and Y is the quotient this problem.

space x * Y = ( X x Y x [ O ) 1])/ - )

where

x = x’ and t = t’ = 0 or { y = y’ and t = t’ = 1.

(z,y,t) is equivalent to (d, y,t’) if and only if

(a) So * So and S’ * So are homeomorphic to familiar spaces. What space

(b) Describe X * So for a general space X . are they?

(Indiana) Solution.

(a) So * So is homeomorphic to the unit circle S’, and S’ * So is homeomorphic to the unit sphere S2.

(b) Generally, X * So is homeomorphic to the quotient space X x [0,1]/ - obtained from the cylinder X x [0,1] by collapsing X x (0) and X x (1) to two points p and q respectively.

2105

(a) Define quotient map. (b) Show that if X is compact, Y is Hausdorff and f : X + Y is continuous

(c) Show that i f f satisfies the condition of (b) then f is a quotient map. and onto, then f is a closed map.

(Indiana) Solution.

(a) Let X be a topological space and - be an equivalence relation on X. Define by X / - the space of equivalence classes under -. By the quotient map ?r : X -+ X/ - we mean the map which assigns to x E X the equivalence

86

class containing x. The quotient space X / - may be topologized by defining a subset U c X/ - to be open if and only if r - ' ( U ) is open in X . Under this topology, r becomes a continuous map. More generally, if f : X + Y is a continuous map, there is naturally associated an equivalence relation on X such that z1 - 5 2 if and only if f(x1) = f(z2). If Y is homeomorphic to X / - under the map i : X/ --+ Y and f = i o T then we call f a quotient map.

(b) Let A be a closed subset of X . Since X is compact, A is compact too. It follows from the continuity of f that f ( A ) is a compact subset of Y . Since Y is Hausdorff, f ( A ) is closed in Y . Hence f is a closed map.

(c) Let - be the equivalence relation on X associated to the map f. Denote by [z] the equivalence class containing 2. Then we define a map i : X/ -4 Y by i([x]) = f (x) . Since f is onto, i is obviously a 1 - 1 map. By the result of (b), the quotient space X / - is compact. Hence i is a continuous 1 - 1 map from the compact space X / N to the Hausdorff space Y , and, therefore, is a homeomorphism. It is clear that f = i o T . So f is a quotient map.

2106

Let f : X ---f Y be a continuous function from a space X to a Hausdorff space Y . Let C be a closed subspace of Y , and let U be an open neighborhood of f-'(C) in X.

(a) Prove that if X is compact then there is an open neighborhood V of C in Y such that f-'(V) c U .

(b) Give a counterexample to show that if X is not compact, then there need not be such a neighborhood V .

Solution. (a) Let W = X - U , then W is closed in X . Since X is compact, W

is compact too. Since f is a continuous function, f (W) is a compact set of Y , and consequently is a closed set of Y because Y is a Hausdorff space. Let V = Y - f (W). Then V is an open neighborhood of C in Y . Since W c f-'(f(W)), we see that

(Indiana)

f-'(v) = x - f-l(f(w)) c x - w = u. (b) Let X = R and Y = S1. f : X -+ Y is the continuous function defined

by j ( t ) = earit for t E R. Take C = (1) E S1. It is obvious that

f-yc) = {n I 72 EX}.

87

Let U = U Un be the open neighborhood of f - ' (C) in X , where n €I

1 1 n n

U , = ( - - + n , n + - ) .

Since 2

n - w n lim lUnl = - = 0,

one can easily prove that there does not exist such a neighborhood V.

2107

Let X be a normal topological space and A c X a closed subspace. (a) Show that the quotient space Y obtained by collapsing A t o a point is

(b) Does this result hold if normality is replaced with regularity? normal.

(Indiana) Solution.

(a) Let R : X ----f Y be the identification map and yo E Y be the point which A collapses to. Let U and V be two nonernpty closed sets in Y such that U n V = 0. Then R- ' (U) and R-'(V) are two nonempty disjoint closed sets in X . If yo $ U U V, then, by the normality of X, it is clear that there exist two disjoint open sets W1 and Wz in X containing R- ' (U) and R-'(V) respectively such that WinA = 8 for i equal to 1 and 2. Thus we see that R ( W ~ ) and R ( W ~ ) are two disjoint open sets in Y and contain U and V respectively. If yo E U (or V), we only need to take the sets W1 and Wz without the restrictions that Wi n A = 0.

(b) Let X be a regular space which is not a normal space. I t means that there exist two disjoint closed sets A and B in X such that they cannot be separated by disjoint open sets in X . Then the quotient space Y obtained by collapsing A to a point yo is not regular, because one can prove that the point yo and the closed set R ( B ) in Y cannot be separated by disjoint open sets in Y .

2108

Let p : E --f B be a covering map with E locally path connected and simply connected. Let X be a connected space, let f : X -+ B be a continuous map,

88

and f l , f2 : X -+ E be two lifts of f . Prove that there is a deck transformation g : E + E such that fi = g f 1 .

Solution. Take a point zo E X and let f l ( z 0 ) = eo E E. Then eo E p- l (bo) , where

bo = p(e0) . Let f2(20) = e l . Since p f 1 = p f 2 = f , we see that e l E p-l(b0). By the assumptions p : E -+ B is the universal covering map. Thus there exists a deck transformation g such that g(e0) = e l . Therefore, g f l ( z0 ) = f z ( 2 0 ) .

Let

(Indiana)

A = {. E x I s f 1 ( z ) = f 2 ( z ) } .

It is obvious that A is not empty. It is not difficult to prove that A is both-open-and-closed in X. Thus, by

the connectedness of X , we see that A = X, which means fi = g f l : X -+ E.

2109

- Let p : X -+ X be an n-sheeted covering projection, n < 00. Suppose that

X is compact. Prove that 2 is compact.

Solution. Suppose that U = {UA,X E A} is an open covering of X . For any point

z E X, let p - ' ( z ) = (21,. .. , Z n } . Let W ( z ) be an elementary neighborhood, i.e., W(x) is an path-connected open neighborhood of x such that each path component of p - ' ( W ( z ) ) is mapped topologically onto W ( z ) by p. Let

(In dian a)

-

where wi(x) is a path component of p - ' ( W ( z ) ) such that Zi E w,(z). Choose a U, E U such that Z , E Ui. Let c ( z ) be the path component of %,(z) n Ui containing Zi. It is obvious that each &(x) is open and R(z) n (x) = 0 for

i # j . Since p is an open map, fi p ( c ( z ) ) is an open neighborhood of z in X .

Choose another elementary neighborhood V(z) of z such that

-

i=l

n

v ( ~ c ) c np(w) . i=l

89

Let n

P-'(v(zc>) = (J W), i = l

where c(z) is the path component ofp-l(V(z)) for any i. Then it is easy to see

that T(z) c E(z) c Ui for i = l , - - + , n . It follows that p-'(V(z)) c 6 Ui.

That is, we have proved that for any point z E X there exists an elementary neighborhood V(z) of 3: such that p-l(V(z)) can be covered by afinite number of sets in U. Since X is compact, X can be covered by a finite number of V(zi), i = 1,. . - , m, and consequently X can be covered by a finite subcover of U.

i = l

-

2110

Let 7 and U be two different topology on X such that X is compact and Hausdorff with respect to both. Prove that 7 p U. (Recall that 7 c U means that every set in the topology 7 is contained in U.)

Solution. We use the reduction to absurdity. Suppose that 7 c U. Let ( X , 7 ) and

( X , U ) denote the topological spaces of X with respect to 7 and U respectively. h : (X,U) -+ ( X , ' T ) is the identity map of X. Then h is a 1 - 1 map from the compact space ( X , U ) to the Hausdorff space (X, 7). We claim that h is a continuous map. For any point 20 E X . Let U be any open neighborhood of 3:o in ( X , 7 ) . Since 7 c U, U is also an open neighborhood of z o in ( X , U ) . It is obvious that h(z0) = zo and h ( U ) = U . Hence h is continuous at 20.

Thus h is a homeomorphism from (X, U) to (X, I), which means U = 1. This contradicts the assumption.

(Indiana)

2111

Let X be a topological space and let A c X. Show that if C is a connected subset of X that intersects both A and X - A , then C intersects BdA. (Recall that BdA = 2 n (X - A ) . )

Solution. We use the reductio ad absurdum. Suppose that C n BdA = 0. Take

U = Cnx and V = C n ( X - A ) . Since C n A c U and C n ( X - A ) c V ,

(Indiana)

from the assumption, we see that both U and V are nonempty subset of C. It is clear that C = U U V and both U and V are closed subsets of C , and, consequently, that both U and V are open sets of C. But

u n v = c n A n ( X - A ) = C n B ~ A = 0,

which is a contradiction to the connectedness of C.

2112

Let ( X , d ) be a metric space. For any subspace A c X and real number E > 0, let

O,(A) C, (A)

= =

{z E A : d ( z , a ) < E for some a E A } , {z E A : d ( z , a ) 5 E for some a E A } .

(a) Prove that O,(A) is an open subspace of X . (b) If A is compact, show that C, ( A ) is closed in X . Must C, ( A ) be closed

for a general subspace A of X ?

Solution. (a) Let 20 be any point of O,(A) . By the definition of O,(A) , there exists

a point a E A such that d(z0, u ) < E , i.e., 6 = E - d(z0, a ) > 0. Then, for any

(Indiana)

2 E 06 /4 (z0 ) ,

6 4 d ( z , a ) ~ d ( z , z o ) + d ( z o , a ) < - + ( E - d ) < E ,

which means that 06/4(zo) c O,(A) . Thus O,(A) is an open subspace of X . (b) Let zo be any cluster point of Ce(A) . Then there exists a sequence

{z,} in C, (A) such that z, # zo for any n and 2, --+ zo as n + 00. By the definition of CE(A) , for each z, there exists an a, E A such that d(z,, a,) 5 E.

Since A is compact, without loss of generality, we may assume that a, -+ a for some a E A. Thus we have

which means 20 E C,(A). So C, (A) is closed in X . If A is not compact, we give a counterexample as follows. Take X = [0,2] c R and A = [0 , 1). Then C I ( A ) = [0, g) is not closed in X .

a

91

2113

Suppose that X is a dense subspace of a topological space Y. Prove or give

(a) If X is Hausdorff, then Y is Hausdorff. (b) If X is connected, then Y is connected.

counterexamples to the following assertions:

(Indiana) Solution.

Y = {a , b, c}. The topology on Y is determined by the family of open sets a) This assertion is not correct. We give a counterexample as follows. Let

7 = { { a , b, e l , {a , C l , { b , c ) , { c } , 0).

Since any neighborhood of the point a always contains the point c, Y is not a Hausdorff space. But it is easy to see that the subspace {c} is dense in Y and is Hausdorff.

b) This assertion is true. We give a proof to it as follows. If Y were not connected, then there would exist a nonempty proper subset U of Y which is both-open-and-closed in Y. Let A = X n U . Since X is dense in Y and U is open in Y, A would be a nonempty open set of X. On the other hand, X - A = X n (Y - U ) would be an open set of X , because Y - U is open in Y . Thus A would be a nonempty subset of X, which is both-open-and-closed in X. Since X is dense in Y and Y - U is open in Y , X - A is nonempty. It means A # X . This contracts the connectedness of X .

2114

Let X and Y be topological spaces, X = U U V , and f : X -+ Y be a

a) If U and V are open in X, show that f is continuous. b) Give an example where U and V are not open in X and f is not con-

function so that flu and flv are continuous.

tinuous.

Solution. a) Let N be an open set of Y. Since f lu and fv are continuous, f - ' Iu(N)

and f - ' l v ( N ) are open in, respectively, U and V . But U and V are open in X , therefore, f - ' I u ( N ) and f - l I v ( N ) are also open in X. Thus

(Indiana)

fF1(N) = f-lIv(N) uf-'Iv(N)

92

is open in X , and consequently, f is continuous. b) Let X = [0,2], U = [0 , 1) and V = [l, 21. f : X + R is defined by

x X E U , f(x) = { 2 XEV.

Then f (u and flv are continuous, but f is not continuous.

2115

Let q : X --t Y be a quotient space projection from a topological space X to a connected topological space Y. Assume that q-l(y) is connected for each y E Y .

a) Show that X is connected. b) Is X necessarily connected if the map q is not assumed to be a quotient

mapping? Justify your assertion. ( Co turn bia)

Solution. a) Suppose that X is not connected. Thus, there exist two disjoint non-

empty open subsets of X , U and V such that X = UUV. Then q(U)nq(V) = 0. For otherwise, let y E q ( U ) n q(V) . Therefore,

q-YY) = (q-l(y) n U ) u (q-l(y) n V ) ,

and it is obvious that q-l(y) n U and q-l(y) n V are both nonempty open subsets of q-l(y), which contradicts the connectedness of q-'(y). Since q is a quotient mapping, V = q- l (q (V) ) and U = q-l(q(U)), q ( U ) and q(V) are disjoint nonempty open subsets such that Y = q ( U ) U q(V) , which contradicts the connectedness of Y . Thus X must be connected.

b) The following example shows that the assumption that q is a quotient mapping is necessary for X to be connected. Let X = U U V , where

U = { ( x , O ) E R2 10 5 2 5 1)

and v = ((1,y) E R2 11 5 y 5 2).

The topology of X is induced from the topology of R2. Let Y = [0,1], the unit interval of R, and : X --+ Y be the map defined by q ( x , 0) = x for ( x , O ) E U and q(1 ,y) = 1 for (1,y) E V . Then q is continuous but is not a quotient mapping. For each y E Y, q- l (y) is connected. But X is not connected.

93

2116

Let X and Y be topological spaces, with Y compact. Let p : X x Y -+ X be the usual projection onto the first factor. Show that p is a closed map.

( Cincinnati) Solution.

Let U c X x Y be a closed subset and zo E X - p ( U). Then, for any y E Y , ( 2 0 , y) 6 U . Since U is closed, there exist an open set Wz,(y) of X and open set V,(ZO) of Y such that (ZO,Y) E Wz0(y) x V,(ZO) and (Wzo(y) x V,(zo))nU = 0. Since Y is compact, there must exist a finite number of V,, (zo), ' . , VYn (zo) such that

n

y = u VYi(XO).

w ( ~ ~ ) = n w ~ , ( ~ ~ ) .

i = l

n Let

i = l

Then W(z0) is an open neighborhood of zo in X . Since (W(z0) x Y ) n U = 0, we see that W(z0) n p ( U ) = 0, i.e., W(z0) C X - p ( U ) . Thus X - p ( U ) is an open set of X , and consequently, p ( V ) is closed in X , which means that p is a closed map.

2117

Let Y be a connected subset of the topological space, and let 2 be a set such that Y is a subset of 2 and 2 is a subset of the closure of Y . Prove that 2 is connected.

(Minnesotu) Solution.

According to the assumptions, we have Y c 2 c F. Let f : 2 -+ So be any continuous map, where So = {-1,1} is the O-dimensional sphere with discrete topology. Since Y is connected, without loss of generality, we may assume that f ( Y ) = {l}, i.e., Y c f-'(l). Taking closures of these two sets with respect to 2 and noting that f-'(l) is a closed subset of 2, we get (Y)z c f-'(l). But as well-known, (Y)z = Y n 2 = 2. Thus we have 2 c f-'(l), and it follows that f is not surjective. It means that there does not exist any continuous surjective map from 2 to So and therefore 2 is connected.

-

94

2118

Let A be a connected subspace of a connected set X . If C is a component of X\A, show that X\C is connected.


We first prove that if X is a connected space, U is a connected subset of X , and if V is a both-open-and-closed subset with respect to X\U, then U U V is connected. Suppose that U U V is not connected. Then let U U V = W1 U W2

where W1 and W2 are two disjoint nonempty both-open-and-closed subsets of U U V. Since U is connected, without loss of generality, we may assume that U c W,. Thus W1 is a both-open-and-closed subset of V , and, consequently, a both-open-and-closed subset of X\U. Hence W1 is a nonempty both-open- and-closed subset of (U U V ) U (X\U) = X , which contradicts the assumption that X is connected. So the above statement is proved.

Now suppose that X\C is not connected. Let X\C = U U V where U and V are two disjoint nonempty both-open-and-closed subsets of X\C. Since A c X\C and A is connected, we may assume that A C V. By the above fact, C u U is connected because C is connected. Since C U U c X\A and CnU = 0, we see that C U U is a connected subset of X\A containing C , which contradicts that C is a component of X\A. Hence X\C must be connected.

2119

1) A metric space X has property S if for every E > 0 there is a cover of

a) Prove a metric space X has property S if X has a dense subset with

b) Suppose X is a subset of a metric space. Suppose the closure of X has

X by connected sets each of which has diameter < E .

property S.

property S. Must X have property S?

Solution. a) Suppose that X has a dense subset A which has property S. Then,

for any E > 0, there is a cover {Val& E I?} of A by connected sets such that each V, has diameter < ~ / 2 . Since each V, is connected, va is also connected

( Cincinnati)

95

and obviously has diameter < E . Since = U Val we see that

{Val Q: E I'} is a cover of X by connected sets each of which has diameter < E.

Thus X has property S. b) We give a counterexample as follows. Let R denote the euclidean real

line, X be the set of all rational numbers. Then x = R has property S, but it is easy to see that X does not have property S.

= X and aEr

2120

Let X be the topologist's sine curve defined by

X = {(z,sin.lr/z) 10 < z 5 1) u ((0,y) 1-1 5 y 5 2) u{(z12) I 5 5 '1 { ( l l Y ) 7 O 5 y 5 2, c R2.

(i) Sketch X . (ii) Let f : X -+ X be continuous. Show that either f ( X ) = X or else there exists 6 > 0 such that

3 3 2 - 2

-- < y 5 -} = 0.

(Toronto) Solution.

(i) X is shown in the Figure below.

Fig.2.1

(ii) Let A = f ( X ) n { ( z , s i n r / z ) 10 < z 5 l}

and 6 = inf{z 1 (z7sin.rr/z) E A } .

96

I f 6 = 0, we claim that f ( X ) = X . To prove it, we first note that since X is path connected, f ( X ) is also path connected. Hence in this case there exists a 60, 0 < 60 5 1, such that

Therefore it is easy to prove that the set

{(O,Y) I -1 I Y I 1) c f (X)*

Once again, using the fact that f ( X ) is path connected, we see that f ( X ) = X. If 6 > 0, then it is obvious that

3 3 2 - 2

f(x) n {(z,Y) 10 < 2 < 6,-- < y 5 -} = 0.

2121

Let T = S1 x S1 denote the torus. (i) Show that T can be covered by 3 contractible open subsets. (ii) Show that T cannot be covered by 2 contractible open subsets.

( Toronto) Solution.

(i) I t is well-known that the torus T can be identified to the quotient space of a square X obtained by identifying opposite sides of the square X according to the directions indicated by the arrows as shown in the Figure below.

n

a

Fig. 2.2

Thus let U1 = X - {u U b } , UZ = X - I and U3 = X - 11. (See the Figure above.) Then U1, U2 and U3 are contractible open subsets and T = U~uU,UU3.

97

(ii) Suppose that T could be covered by 2 contractible open subsets. From the Van Kampen theorem it would follows that the fundamental group ?rl(T) = 0. It contradicts the known fact that a,(T) x Z @Z.

2122

Let U = (U,}aE~ be an open cover of the space X. a) Give the definition for “U is locally finite”. b) If U is locally finite show that, for any subset K c J, U r p is closed.

P E K ( Toronto)

Solution.

neighborhood which intersects only a finite number of U,. a) By definition, U is said to be locally finite, if each point p E X has a

b) For any point p $! u vp, since U is locally finite, there is an open P E K

neighborhood W of p which intersects only a finite number of sets in U, particularly, only a finite number of Up for /3 E K , say, Up, , . . . , Up,. Since p $! r p ; for each pi, there is an open neighborhood V , of p such that V, nupi = 0. Then let V = W n V1 n . . . n V,. It is clear that V is an open neighborhood of p such that V n ( u up) = 0. Hence u vp is closed.

PEK P € K

2123

Let S be a set and let F be a family of real valued functions on S such that f ( s 1 ) = f ( s 2 ) for all f E F implies s1 = s2. Prove that there exists a weakest topology in S amongst all those for which all members of F are continuous. Show further that the resulting topological space satisfies the Hausdorff separation axiom.

Solution. Let

( H a m a 4

U = { f - ’ ( ( a , b ) ) 1 (a , b) is any open interval of R and f E F } .

Then there exists a unique topology 7 on the set S of which U is the topology subbase. It is easy to see that 7 is the weakest topology on S amongst all those for which all members of F are continuous. Suppose that s1 and s2 are

98

two distinct points of S. By the assumption there exists a t least an f E F such that f(s1) # ~ ( s z ) . Hence we may take two open intervals (a1,bl) and ( a z , b z ) such that f ( S i ) E (ai ,bi) for i = 1 , 2 and (a1,bl) n ( U Z , bz) = 0. Then Ui = f-'((al, bl)) and Uz = f - l ( ( a z , b z ) ) are two disjoint open sets of (S, 7) such that si E U, for i = 1,2. So (S, 7) is Hausdorff.

99

SECTION 2 HOMOTOPY THEORY

2201

a) Give generators and relations for the fundamental groups of the torus

b) Compute the fundamental group of the figure 8 and draw a piece of its and of the oriented surface of genus 2.

universal covering space.

Solution. a>

As is well-known, we can present the torus T as the space obtained by identifying the opposite sides of a square, as shown in Fig.2.3 (b). Under the identification the sides a and b each become circles which intersect in the point 20. Let y be the center point of the square, and let U = T - {y}. Let V be the image of the interior of the square under the identification. Since V is simply connected, by the Van Kampen theorem, we conclude that ?rl(T, TI) is isomorphic to 7~1(U, z1) modulo the smallest normal subgroup of ?rl(U, 21)

containing the image +*(rl(UnV, TI)), where qL is the homomorphism induced

100

by the inclusion map q5 : U n V -+ U . It is easily seen that a u b is a deformation retract of U . Hence a l ( U , 20) is a free group on two generators a and p, where a and p are presented by circles a and b, respectively. It is also clear that T ~ ( U , Z ~ ) is a free group on two generators a’ = F1a6 and p’ = C1p6, where 6 is the equivalence class of a path d from 20 to 21. (See Fig.2.3 (b).) On the other hand, it is easy to see that x l ( U n V,zl) is an infinite cyclic group generated by 7, the equivalence class of a closed path c which circles around the point y once, and, consequently, that q5+(y) = c y ’ p I ~ y ’ - ~ f l ‘ - ~ . The smallest normal subgroup of a l (U , 21) containing &(al (U n V, 21)) is just the commutator subgroup of 7rl(U, 21). Thus ?rl(T, 21) is a free abelian group on two generators a’ and p’. Changing to the base point 20, we see that rl(T, 20) is a free abelian group on two generators a and f l , which are presented by circles a and b, respectively.

In a similar way, we can see that the fundamental group of the oriented surface of genus 2 is a free group on four generators a1,p1, a2,p2 with the single relation [a1,/31][a2,p2], where a l , p 1 , a2 , /32 are presented by circles a l , bl,aa,ba,

respectively, and [ail pi] denotes the commutor aipiailpzrl. (See Fig.2.3 (a).) b)

Fig .2.4

Let X denote the figure 8 space, as shown in Fig.2.4. Let U = X - { q } , V = X - { p } . By the Van Kampen theorem we can prove that a l ( X , z o ) is a free group on two generators a,p, where a ,P are presented by circles a and b, respectively.

The following picture is a piece of its universal covering space.

it

Fig .2.5

Under the covering map T , each level segment is mapped on the circle a

101

according to the direction indicated by the arrow, and each vertical segment is mapped on the circle b according to the direction indicated by the double arrows.

2202

Let A be a connected, closed subspace of a compact Hausdorff space X , and suppose f : A -+ A is a continuous map. For each positive integer n let fn(A) = ~ o ~ o - . . o ~ ( A ) . -

n times 00

(i) Show that B = n f"(A) is connected.

ii) Suppose y : S1 - X - B is a nullhomotopic map. Show that there exists a positive integer n such that y(S1) c X - f"(A) and such that the induced map y' : S1 + X - fn(A) (y'(s) = y(s) for s E S') is also nullhomotopic.

Solution. i) Since X is compact and A is closed, A is also compact. Thus it is easy

to see that f"(A) is compact, closed and connected. Noting that ,,+'(A) c f"(A) c A for any n, and that A is compact, we see that B = n fn(A)

is a nonempty closed subset of A. Suppose that B is not connected. Then B = U u V , where U and V are disjoint nonempty closed subsets of B. It is obvious that U and V are also closed subsets of A. Since A is obviously compact and Hausdorff, A is a normal space. Therefore, there exist disjoint open subsets of A, Wl and Wz such that U c W1 and V c W2. Thus it follows that B c W1 U Wz = W , which is an open subset of A. We claim that there exists a positive integer N such that fN(A) c W . For, otherwise, there would be a squence in A , {xn}, such that x, E f"(A) but 2, @ W for every n. It means that x, E A - W for every n. Since A - W is closed in A and, consequently, is a compact subset, there exists at least a limit point 20 of the sequence { x C n } . Noting that fn(A) is compact for every n and fn+'(A) c fn(A), we can see that xo E B c W , which is a contradiction. Thus f N ( A ) c W , and therefore,

n=l

(Indiana)

m

n = l

P Y A ) = ( f N W n Wl) u (fN(A) n W2>,

which contradicts the fact that fN(A) is connected. ii) Let F : S1 x I 3 X - B be a homotopy between y and the constant

map. Since F(S1 x I) is compact and X - B is open, F(S' x I) is also closed in

102

X . Since X is normal, there exists open sets U and V such that F(S1 x I ) c U and B c V . From the proof for i), it follows that there exists a positive N such that fN(A) c V . Therefore,

F(S1 x I ) c u c x - fN(A). Particularly, y(S') c X - fN(A). The remainder of ii) is obvious.

2203

Let RP" be the real projective n-space and T" be the rn-torus S1 x . . - x S1 ( m factors). Prove that any continuous map RP" -+ Tm is null-homotopic.

Solution. It is well-known that R" is a universal covering space of T". Let P :

R" -+ T" be the universal covering map. It is also well-known that

(Indiana)

ai(T") M Z e . . . (rn times)

and that al (RPn) M Z2. Therefore, for any continuous map f : RP" -+ T", the induced homomorphism - f+ : a1(RT) -+ nl(T") is trivial. Thus there exists a lifting of f , say f, such that pf = f . Let po be a fixed point of R". Define H : RP" x I -+ R" by

H ( z , t ) (1 - t).T(z) + tpo. -

Then H is a homotopy between f and the constant map PO. So fl is nullhomotopic and consequently f is also null-homotopic.

2204

Let X be the quotient space obtained by collapsing { p t . } x S1 c S1 x S1 to a point

Compute x1(X) and H , ( X ) .

Solution.

103

(Columbia)

a

F ig.2.6

X may be identified with the space shown in Fig.2.6, which is obtained by identifying the sides of a 2-gon. Let y be the center of the 2-gon, U = X - {y}, and V be the interior of the 2-gon. Then, U and V are open subsets, U , V, and U n V are path connected, and V is simply connected. Thus, by the Van Kampen theorem, 7r l (X , z l ) is isomorphic to the quotient group of a l (U, 21) modulo the smallest normal subgroup of R ~ ( U , 21) containing the image q5,(al(U n V, zl)), where q5* is the homomorphism induced by the inclusion map q5 : U n V -+ U . It is easy to see that u is a deformation ratract of U . Therefore, al(U, zl) is an infinite cyclic group on the generator S-luS, where S is the equivalence class of a path d connecting 20 and z1. (See Fig.2.6) It is also clear that x l ( U n V, zl) is an infinite cyclic group on the generator y, where y is the equivalence class of a closed path c which goes once round the point y. It is easy to see that &(y) = lrl(u,zl). Therefore we conclude that

To compute H , ( X ) , we may apply the Mayer-Vietoris sequence to the pair ( U , V). The conclusion is that H i ( X ) is an infinite cyclic group for i equal to 0 , 1 , 2 and is zero otherwise.

.l(X) =z.

2205

(i) Suppose n 2 2. Does there exist a continuous map f : S" + S1 which

(ii) Suppose n 2 2. Does there exist a continuous map f : RP" --+ S1

(iii) Let T = S1 x S1 be the torus. Does there exist a continuous map

is not homotopic to a constant?

which is not homotopic to a constant?

f : T -+ S1 which is not homotopic to a constant? ( Toronto)

104

Solution. (i) Let a : R -+ S1 be the universal covering map defined by ~ ( t ) = eZrrit ,

and f : S” --+ S1 be a continuous map. Since al(S”) = 0 for n 2 2, we see that there is a lifting of f, 7 : S” --+ R such that T? = f. Since R is contractible, ?must be homotopic to a constant map : S” -+ R, hence f is homotopic to a o c = C, which is also a constant map from S” to S1, i.e., there does not exist any continuous map f : Sn -+ S1 which is not homotopic to a constant map.

(ii) Since al(RPn) = Zz, any continuous map f : RPn -+ S1 induces a trivial homomorphism f* : x1(Rrn) -+ al (S1) , and, consequently, has a lifting 7: RP” -+ R such that a o f = f. By the same argument as in (i), f must be homotopic to a constant.

(iii) Denote T = S1 x S1 by T = (e2rr i t l , e2ni t2 ), 0 5 t l , t z 5 1. Define f : T + S 1 by

f ( p i t l , e 2 x i t z ) = e27ritl E sl. It is easy to see that the induced homomorphism f* : H1(T) -+ H1(S1) maps one generator of H1(T) to the generator of H1(S1), and another generator to zero. Hence f* is not trivial, which means that f is not homotopic to a constant.

2206

A continuous map of topological spaces: p : E -+ B is called a fibration if it has the homotopy lifting property - that is, for any pair of continuous maps I --t G : X x I -+ B and h : X x (0) --t E such that p h = Glxx(oi there exists a continuous map H : X x I --+ E such that H l x x i o ) = h and p H = G. Let p : E -+ B be a fibration, bo E B be a base point, F = p - l ( b o ) (the “fiber”), and eo E F . Let i : F --+ E denote the inclusion map.

(i) If F is path connected, prove that p# : ?rl(E, eo) -+ .rrl(B, bo) is surjective.

(ii) Prove that in general the 3-term sequence

Tl(F, eo) + Tl(E, eo) -+ “1(B, bo)

in which the homomorphisms are i# and p # , respectively, is exact.

Solution. (i) Let cr = [f] E ~ l ( B , b o ) , where f : I -+ B is a closed path at bo

which represents a. Let G : I x I --f B be a continuous map defined by

(Indiana)

105

G(t,s) = f ( s t ) for ( s , t ) E I x I , and h : I x (0) --+ E be the constant map such that h(t,O) eo. It is obvious that p h = GII,{o}. Therefore there exists a continuous map H : I x I + E such that H / I x { o } = h and p H = G . Particularly we have p H ( t , 1) = f ( t ) . Let C : I -+ E be a path in E defined by c ( t ) = H ( t , 1). Then pc = f . Let c ( 0 ) = e l and c(1) = e2. It is clear that el and e2 belong to F. Since F is path-connected, we may choose two paths g1 and 92 in F such that gl(0) = e2, gl(1) = e l , g2(0) = el and gz(1) = e l . Thus f = g2 * c * g1* gF1 is a closed path at e l , where gF1 is the inverse path of g2. Noting that pgl, pg2 and pgz' are all constant path at bo, we have

-

P# rfi = bfi = bg21 . b.1 . bg11 . h , l I = bl = [fl = a,

which means that p# is surjective. (ii) Let a = [f] E .Irl(F,eo). It is obvious that p# . i#(a) = [pf] = [bo],

where bo is the constant - path at bo. Hence im, c kerp#. On the other hand, suppose that G = [f] E kerp#. Then there exists a homotopy G : I x I --+ B between pTand the constant path bo such that G(t, 0) = ( p T ) ( t ) , G(t, 0) = bo, and G(0, s ) G(l , s ) E bo for any s. By the homotopy lifting property, there - exists a continuous map H : I x I --+ E such that P H = G and H ( t , 0) = f(t). Let c1 = HI{o ixr , cg = H11,{1} and c3 = H l { l } x ~ . Then c1, cg and c3 are paths in F and cl(0) = eo, ~ ( 1 ) = cz(O), ~ ( 1 ) = ci1(0) and c i 1 ( l ) = eo. Hence c = c1 x c2 x c i l is a closed path in F with base point eo. It is easy to see that f is homotopic to C. (See Fig.2.7.) Therefore, G = [fi = [c] = i#[c]. i.e., kerp# c imi#. Hence the sequence mentioned above is exact.

Fig.2.7

2207

Is the canonical map q : S2 --+ RP2 (which identifies antipodal points of S2) nullhomotopic? Why or why not?

(Indiana)

106

Solution. It is well-known that q is also the universal covering map from S2 to RP2.

Suppose that q is nullhomotopic. Then q is homotopic to a constant map c : S2 --+ RP2, and therefore q has a lifting < : S2 -+ S2 which is also homotopic to the lifting of c , a constant map C : S2 + S 2 . But it is obvious that f is the identity map or the antipodal map from S2 to S2. Hence degg = fl . On the other hand, we have d e g c = degZ = 0. This is a contradiction. Thus we conclude that q is not nullhomotopic.

2208

Let p : E ---f B be an universal cover with E and B path-connected and locally path-connected. Let T : B + B be a map so that Tn = Id and so that T(b) = bfor some b E B. (Here Tn = T o T o - - . o T , n times.) Show that there is a map ? : E --+ E so that p o T = T o p and Tn = Id.

Solution. We have

T o p(eo ) = T ( b ) = b. Since ?rl(E) is trivial, there is a lifting of T o p , T : ( E , eo) + ( E , eo) such that p o ? = T o p. Since

- - (Indiana)

Choose eo E p- l (b ) and consider the map T o p : E + B . I

- - - - T" = ( p T ) ~ " - l = T ( p ~ n - l ) = . . . = T~ O P

- - - - - and T"(e0 ) = T"- ' (T(eo) ) = T*-'(eo) = ... = T ( e 0 ) = eo,

we see that @ is a lifting of T" o p at the base point eo. On the other hand, since T" = Id, it follows that identity map Id : (E, eo) + ( E , eo) is obviously a lifting of T n p a t the base point eo. Therefore, by the uniqueness of lifting, we conclude that r?;. = Id.

2209

Let T 2 = S1 x S1, and let X c T2 be the subset S1 x { 1) U (1) x S1. Prove that there is no retraction of T 2 to X .

Solution. We use the reduction to absurdity. Suppose that there is a retraction of T2

to X , denoted by r. Let i : X --t T2 be the inclusion map. Then roi : X --+ X is

(Indiana)

107

the identity map. Hence ~,oi, : r l ( X ) -+ r l ( X ) is the identity homomorphism. It is well-known that r1(T2) is abelian and that .1(X) is an non-abelian free group on two generators denoted by a and b. Hence we have ab # ba and i*(a) i*(b) = z*(b)i*(a). Therefore we have

which is a contradiction.

2210

Let A c R3 be the union of the x and y-axis,

A = {(X,Y,Z) I (Y2 + z2)(x2 + 2) = O},

and let p = ( O , O , 1). a) Compute H1(R3 - A) . b) Prove that rl(R3 - A , p ) is not abelian.

(Indiana) Solution.

Let X = {(x, y, z) I x2 + z 2 = 1) be a circular cylindrical surface. pl and p2 denote the points (1,0,0) and ( - l , O , 0) respectively. Then it is easy to see that X - {pl ,p2) is a deformations retract of R3 - A. It is also clear that X - { p l , p z ) is homotopically equivalent to the space Y as shown in Fig.2.8.

Fig.2.8

a) Y has a structure of a graph with 4 vertices and 6 edges. The Euler Characteristic K ( Y ) = 4 - 6 = -2. Hence the rank of H I ( Y ) is equal to 1 - (-2) = 3. Therefore we conclude that H1(R3 - A ) M H 1 ( Y ) z Y Z @YZ @Z.

b) Let e be a point on the arc ab different from a and b. Take U = Y - { c } and V = Y - { e } . Then we have Y = U U V , It is easy to see that U has the

108

same homotopy type as S1 and that V has the same homotopy type as the “eight figure space”. Since U n V is contractible, by the Van Kampen theorem we see that ?rl(Y) is a free group generated by ?rl(V) and ?rl(V). Therefore we conclude that ?rl(R3 - A , p ) M ?rl(Y) is a free group on three generators and, consequently, that ( R3 - A , p ) is not abelian.

2211

- Let p : (?,g) + (Y,y) be a regular covering space; that is, p*(.rrl(Y,g)

is a normal subgroup of 7rl(Y,y). Suppose that f : X -+ Y is a continuous function from the path-connected space X to Y with f ( z 0 ) = f(z1) = y, and that there is a lifting of f, TO : ( X , ZO) --+ ( Y , 9. Show that there is a second lifting, 5 : (X,z1) + (?,g. (A need not be distinct from 5.)

Solution. (Indiana)

- Let fo(zl) = y. Since

Since p is a regular covering space, there is a deck transformation y such that y(y) = g. Then let - - -

f l = r o f o :x + Y.

Hence 5 is the lifting o f f we want.

2212

Let X be the identification space obtained from a unit 2-disk by identifying points on its boundary if the arc distance between them on the boundary circle is 9. Compute the fundamental group of X.

(Indiana)

109

Solution.

Fig.2.9

Take a point y in the open disk. (See Fig.2.9.) Let U = X - {y} and let V be the open disk. Then both U and V are path connected subsets of X and X = U U V . Since V is simply connected, by the Van Kampen theorem, r 1 ( X ) is isomorphic to the quotient group of r l ( U ) with respect to the least normal subgroup containing 4*(rl(U n V ) ) , where 4 : (U n V ) + rl(U) is the homomorphism induced by the inclusion 4 : U n V -+ U . Take a point z1 E U n V as the base point. (See Fig.2.9.) It is clear that r l ( U n V, zl) is an infinite cyclic group generated by y o the closed path class of a closed path c which circles around the point y once. Since the circle a is a deformation retract of U , it is clear that wl(U, zo ) is an infinite cyclic group generated by u' the closed path class of a. Therefore r l ( U 1 21) is an infinite cyclic group on generator Z = y-'a'y, where y is the path class of a path d from z o to z1. It is also clear that 4+(yc) = 3Z. Hence the least normal subgroup containing 4*(rl(U n V ) ) is isomorphic to 32. It follows that rl(X) NNZG/U =as.

2213

Sketch a proof of the Fundamental Theorem of Algebra (every nonconstant polynomial with complex coefficients has a complex zero) using techniques of algebraic topology.

Solution. Let denote the complex plane and f (2) be a polynomial of positive degree

with complex coeficients. We may consider f to be a continuous nonconstant map f : a' -+a. Note that I f(z)l + 00 as JzI + 00; hence, we may extend f

(Indiana)

110

to a map of the one-point compactification of (X

f : s2 -+ s2 by setting f(m) = 00, where 00 denotes the north pole. Then we may first prove that if f (z ) = zk, k > 0, then the degree of the extension f : S2 -+ S2 is equal to k. Furthermore, we may prove that if f is any polynomial of degree k > 0 then the degree of the extension f : S2 -+ S2 is still equal to k . Noting the fact that if a continuous map f : S2 + S2 is not surjective then the degree o f f is zero, we may prove the Fundamental Theorem of Algebra by means of the reduction to absurdity.

2214

Let X denote the subspace of R3 that is the union of the unit sphere S2, the unit disk D2 in the 2-y plane, and the portion, call it A, of the z axis lying within S2 .

(a) Compute the fundamental group of X . (b) Compute the integral homology groups of X .

(Indiana) Solution.

(4

Fig.2.10

It is clear that X has the homotopy type of the one point union X1 v X 2 where X 1 and X2 are each homeomorphic to the union of the unit sphere and the portion of the z axis lying within S2. (See Fig.2.10.) To compute ~ l ( X 1 v X 2 ) , we take U = X 1 V X2 - { p } and V = X 1 V X2 - (4). Then we have U u V = X1 V Xz . Since U n V is contractible, by the Van Kampen, a l ( X 1 v X 2 ) is a free product of the groups r l ( U ) and T~(V) with respect to the homomorphisms induced by the inclusion maps. It is obvious that

111

Therefore, ? r l ( X ) is a free product generated by two generators. We can take as generators the closed path classes which are determined by the closed paths omp and omq respectively. (See Fig.2.10.)

(b) Applying the Mayer-Vietoris sequence to the pair (V, V), we see that

Noting that Hi(X1) M Hi(X2) which are infinite cyclic for i equal to 0 ,1 ,2 and are zero otherwise, we conclude that

i = 0, otherwise.

Let X be a path-connected space, f : X ---f Y a continuous function, and 20, 21 E X . Suppose that the induced homomorphism

is surjective. Show that

is also surjective.

Solution. Since X is path-connected, there exists a path C : [0,1] ---+ X such that

c(0) = 20 and c(1) = 2 1 . Then Z= f o c is a path connecting f(z0) and f(z1) in Y . For any [a] E ?r I (Y , f ( z l ) ) , where a is a closed path at f ( ~ l ) , cac is a closed path at f(z0). By the assumption, there is a closed path h at 20 such that f*([h]) = [Z?']. It means that f o h and Zu?' are homotopic. Thus f o (c-'hc) and a are homotopic, and, consequently, f*([c-'hc]) = [a]. Hence

(Indiana)

- - - I

f* : m ( X , 2 1 ) n ( Y , f ( 2 1 ) )

is surjective.

112

2216

Let B denote the “figure eight space”. Let p : X -+ B and q : Y -+ B be 2-fold covering maps, where both X and Y are connected. Prove that X and Y are homotopy equivalent, but not necessarily homeomorphic.

Solution. (Indiana)

a m b

Fig.2.11

For any 2-fold covering map p : X -+ B, let p - l ( z 0 ) = {eo,el}. (See Fig.2.11.) Since the automorphism group A ( X , p ) FZ& and X is connected, it is not difficult to see, by considering the liftings of the circles a and b in X, that, in substance, X has only two different types as shown in Fig.2.11. Then it is easy to see that they are homotopy equivalent to an 3-leaved rose G3. But the spaces X and Y shown in Fig.2.11 are not homeomorphic. Otherwise, suppose that f : X -+ Y is a homeomorphism. Then X - { e o ) is homeomorphic to Y - {f(eo)}. Since X - { e o } is contractible and Y - { f ( e O ) } is obviously not contractible, we come to a contradiction.

2217

Calculate the fundamental group of the space RP2 x S2. (Indiana)

Solution. By the formula

.rrl(X x Y) = m(X> @ m(Y>,

we see that

7rl(RP1 x S2) x T1(RP2) @ “1(S2) M a2 @ {O} =a,.

113

2218

Let 2 denote the figure 8 space, Z = X V Y , X and Y circles. Let x, y E TI(.!?, *) be the elements in 2 defined by X , Y , where * denotes the vertex

Fig.2.12

(a) Let h : ~ ' ( 2 , *) -+ Z/62 be the homomorphism satisfying h ( z ) = 2 and h(y) = 3, and let p : 2 -+ 2 denote the covering space corresponding to the kernel of h. ( p * ( ~ 1 ( 2 , *)) = ker(h).) If A is a path component of p - ' ( X ) and B is a path component of p - ' ( Y ) , how many intersection points of A and B are there? (i.e., what is the cardinality of the sat A n B?)

(b) If G is a finite group, h : ~ ' ( 2 , *) -+ G a surjection, and p : 2 -+ 2 the corresponding cover, prove that the number of intersection points of a path component A of p - ' ( X ) with a path component B of p - ' ( Y ) divides the order of G.

Solution. (a) Since h ( z ) = 2 and h(y) = 3 and T ~ ( Z , * ) is generated by 2 and y, it

follows that the homomorphism h is surjective. Thus ~ l ( 2 , *)/ ker h is isomorphic to the group Z/6Z = Zg. Hence the covering space p : 2 -+ 2 is a 6-fold cover. We denote p- ' (*) by

-

-

(Indiana)

p- ' (*) = {eO,el,e2,e3,e4,e5}.

Then, from h ( z ) = 2, we see that the path component of p - ' ( X ) contains exactly the points {eO,ez,e4} of p- ' (*) which corresponds to the elements {0,2,4} of& respectively. By the same reason, we see that the path component of p - ' ( Y ) contains exactly the points {eo, e3} of p-'(*). Since

p-'(x) n p - ' ( Y ) = p - l ( * ) ,

we conclude that A n B = { e o } .

114

(b) Since h is a surjection and G is a finite group, the corresponding cover p : 5 + 2 is a finite fold cover. Suppose that h(z ) = r and h(y) = s. Then r generates a subgroup H1 of G and s generates a subgroup of H2 of G. In a similar way as in (a), we see that the number of intersection points of A and B is equal to the number of elements in H1 n Ha. By Lagrange’s theorem, it divides the order of G.

2219

Let X be the result of attaching a 2-cell 0’ to the circle S1 by the map

(a) Compute, with proof, the fundamental group of X . (b) Compute, with proof, the homology of the universal cover of X .

f : S1 -+ S1 given in terms of complex numbers by z -+ z6.

(Indiana) Solution.

(4

a d @ 20 a

Fig.2.13

Represent X as the space obtained by identifying the edges of a hexagon, as shown in Fig.2.13. Under the identification the edges a become a circle through the point 20. Let y be the center point of the hexagon, and let U = X - {y}. Let V be the image of the interior of the hexagon under the identification. Then, U and V are open subsets, U , V , and U n V are arcwise connected, and V is simply connected. Let 21 be a point in UnV. It is clear that U n V has the same homotopy type with S1, and that wl(U nV, 21) is an infinite cyclic group generated by y, the homotopy class of a closed path c which circles around the point y once (see Fig.2.13.)

Applying the Van Kampen theorem, we conclude that

$1 : .Irl(U, 21) -+ .rrl(X, 21)

is an epimorphism and its kernel is the smallest normal subgroup containing the image of the homomorphism

41 : rl( u n v, zl) + r1(U, zl),

115

where It is obvious that the circle a is a deformation retract of U. Thus ?rl(U, zo)

is an infinite cyclic group generated by a and, consequently, 7r l (U , z l ) is an infinite cyclic group generated by a' = ?-'a?, where 7 is the homotopy class of a path d from zo to 21.

It is obvious that dl(r) = ar6. Hence the smallest normal subgroup containing 41(.lrl(U n V, 21)) is isomorphic t o U. Thus we conclude that

and 41 are homomorphism induced by inclusion maps.

(b) Since x is a finite 2-dimensional cw complex and T ~ ( X ) = its universal covering space X is a 6-fold covering space and is also a 2-dimensional CW complex. I t is well-known that the Euler characteristic X ( 2 ) = 6 X ( X ) . But it is easy to see that X ( X ) = 1, hence X ( 2 ) = 6. From H o ( 2 ) M I , H I ( % ) = 0 and H 2 ( 2 ) x H 2 ( X , X ' ) , where 2' is the 1-skeleton of 2, we see that H 2 ( 2 ) is a free Abelian group of rank 5. So we conclude that

-

- -

( 0, i 2 3,

i = 1, Hi(%) =

I Z, i = 0.

2220

If X is any topological space and S1 denotes the unit circle in the complex plane with its usual topology as a topological group with multiplication given by the multiplication of complex numbers, then it is known that the set [ X , S1] of homotopy classes of maps from X to S1 inherits a natural group structure.

(a) Define this group operation explicitly and indicate the group identity and how inverses are formed. You do not need to prove your assertions.

(b) Compute this group explicitly for X = point, S1, S 2 , and T2 = S1 x S1. (Indiana)

Solution.

s1 as

Then the multiplication of [X, S1] is defined by [f] - [g] = [f . g ] , where the map f . g : X + S1 is defined by ( f a g)(z) = f (z ) g(z) for any z E X . Here

(a) We denote the homotopy class of a map f : X --+ S1 by [f] and write

s1 = { e2= io ECT I o 5 8 5 1).

116

f (z) .g(z) is defined by the multiplication of complex numbers. The identity of this multiplication is [el, where the map e : X + S1 is defined by e ( z ) E eZni for any 3: E X. The inverse of [ f ] is the homotopy class of a map f, which is defined by f(3:) = l / f (z ) for any 3: E X .

(b) If X = point, then [ X , S1] obviously has only one element. So the group [ X , S1] is trivial. For the case of X = S1, one can easily prove that

[S’, S l ] x “l(S1) xa.

For the case of X = S2, it is easy to see that each homotopy class [ f ] can be represented by a map f : S2 -+ S1 which maps the northpole N of S2 to the point po = eZUi of S1. Then by the facts that S2 is simply connected and the universal covering space of S1, R is contractible, one can easily prove that [S2, S1] M r2(S1) = 0. Now we discuss the case of X = T 2 = S1 x S1. For any map f from S1 x S1 -+ S1 we define two maps f1 and f2 from S1 + S1 by fl(6) = f ( e Z n i e , p o ) and fZ(6) = f ( p O l e z U i e ) for any e2nie E S1, respectively. Then let 4 : [X,S1] +Z@Z is defined by 4([f]) = (degf1,degfz). We have

4“fI . [sl) = 4Uf. sl) = (deg(f * 911, deg(f .9)2)

(deg(f1 * Sl) , deg(f2 .92)) (deg fl + degg1, deg f 2 + deg 92)

(deg fll deg f 2 ) + (deg 91 1 deg 92)

= = =

= 4(V1) + N g 1 ) -

Therefore, 4 is a homomorphism from [X, S 1 ] to Z @Z. Note that fl can be extended to a map from X to S1, still denoted by f1, by

which is homotopic to f under the homotopy map F : X x I -+ S1 defined by

1. ~ ( ~ 2 r r i 0 , e2n iG, t ) = f (e2ni f31 e2xit+2xi(l-t)+

Thus one can easily prove that 4 is a monomorphism. It is clear that 4 is a epimorphism, and cosequently q3 is an isomorphism. We conclude that [X,S1] %Z@Z.

117

2221

If X is a path-connected space whose universal cover is compact, show that ?rl(X, 20) is finite.

Solution. Let ?r : 2 -+ X be the universal cover of X. If n l ( X , 20) were not finite,

then ~ ~ ~ ( 2 0 ) would be a closed set of infinite points in X. Since 2 is compact, T - ~ ( z o ) must have at least a limit point, say Z, such that n(Z) = 20. Thus it is easy to see that x is not a local homeomorphism at Z, which is a contradiction.

(Indiana)

-

Prove that if X is locally path connected and simply connected then every map X 4 S1 is homotopic to a constant. What can you say if we just assume that X is path connected, locally path connected and the fundamental group of X is finite?

Solution. Let exp : R -+ 6 denote the exponential covering map, i.e., the universal

covering space of S1. Since X is locally path connected and simply connected, 7r1(X) = 0, and f*(?rlT1(X)) = 0 for any map f : X -+ S1. Hence there exists a lifting of f, ?: X -+ R such that exp(7) = f . Since R is simply connected, f is homotopic to a constant map Z. Denote the homotopy between 7 and E by g. Then exp(2) is the homotopy between f and c, i.e., f is homotopic to a const ant map.

Suppose that ?rl(X) is finite. Since ?rl(S1) % Z and f * ( ~ l ( X ) ) is a finite subgroup of ?rl(S1), we see that f * ( ~ l ( X ) ) must be trivial. So the above argument still works in this case, and the same conclusion holds.

(Indiana)

-

118

SECTION 3 HOMOLOGY THEORY

2301

Prove the 3 x 3 Lemma. Consider the following commutative diagram of abelian groups

0 1

0 ---t A3

62 1 0 -+ B3

61 1 0 -+ c3

1 0

If all 3 columns and the first two rows are short exact, then the last row is also short exact.

Solution. To prove the exactness at C,, we show that 7 2 is injective. Let c E C3 and

7 2 ( ~ ) = 0. Since b1 is surjective, there is a b E B3 such that c = b1(b). By the commutativity we see ,&(b) E kercl. Hence there is a n a2 E A2 such that ~ ~ ( a 2 ) = Pz(b ) . Then since

(Hamud)

c2(al(a2)) = Pl(EZ(a2)) = Pl(PZ(b)) = 0

and C2 is injective, we have al(a2) = 0, i.e., a2 E ker 01 = imcyz. Thus there is an a E A3 such that az(a) = a2. Since

P2(62(a) - b ) = P262(a) - h ( b ) = w 2 ( a ) - Pz(b) = E z ( a 2 ) - Pz(b) = 0

and f ~ 2 is injective, we see &(a) = b. Therefore

c = S,(b) = 6162(u) = 0.

Hence 7 2 is injective.

119

Now we prove that ker y1 c imy2. For any c E ker 7 1 , since ~1 is surjective, there is an b E B2 such that c = ~ l ( b ) . Thus it is easy to see that Pl(b) E ker C1, and consequently that there is an a1 E A1 such that &(u1) = Pl(b). Since a1 is surjective, there is an a2 E A2 such that a l (u2) = ul . Thus by the commutativity, we have b - ~ 2 ( u 2 ) E kerpl. Therefore there is a b3 E B3 such that p 2 ( b 3 ) = b - ~2(u2) . Then

7 2 ( 6 1 ( b 3 ) ) = E l h ( b 3 ) = &l(b) - E 1 & 2 ( a 2 ) = c.

Hence c E imy2. In a similar way we may prove that imy2 c ker 71. Thus the exactness a t C2 is proved.

We leave the proof of the exactness a t C1 to the reader.

2302

Prove that if M is a compact manifold of odd dimension, then X ( M ) = 0. Show examples of compact 4-manifolds with X = 0,1,2,3,4.

Solution.

Therefore,

( Columbia)

Since M is compact, M is 272-orientable. Suppose that d i m M = 2m + 1.

2m+ 1

X ( M ) = (-l)i dimHi(M,&). i=O

By the PoincarC duality theorem,

for any i. Thus, since i and 2m + 1 - i have different parity, they appear in the sum with opposite signs. Therefore X ( M ) = 0.

Let Xo = T2 x T2, X 1 = R P 2 x R P 2 . Denote by uh the connected sum of h projective planes. Then it is well-known that X ( U h ) = 2-h. Let X2 = U3 x 174,

X3 = U3 x Us, and X4 = S2 x S2. Using the fact that

we see that X(X,) = i for i equal to 0 ,1 ,2 , 3 and 4.

120

2303

Let X be a topological space. The suspension E X of X is defined to be the identification space obtained from X x [-1,1] by identifying X x (-1) to a point and X x { 1) to another point. For example the sphere S" is the suspension of S"-' with the north and south poles corresponding to the two identification points.

Compute the homology of E X in terms of the homology of X . (Illinois)

Solution. Let pl and pa be the identification point respectively. Set U = E X - {pl}

and V = C X - { p z } . Then U and V are open sets of E X , and E X = U UV. It is obviously to see that U and V are contractible spaces and X is a deformation retractor of U n V . By Mayer-Vietoris sequence, we have

- where H q - l ( X ) denotes the reduced homology of X

2304

Consider the chain complex

c = {C" In>O} C" =z3 d, : C, --t C,-l defined by

( T , s , ~ ) + (s - t , O , O ) n even

( T , s , ~ ) + ( O , s + t , s + t ) nodd .

Compute H,(C) for all n.

Solution.

en}, i.e., Z, (c ) is isomorphic to Z $Z.

even n.

(Illinois)

When n is even, ( T , s , ~ ) E Zn(c) if and only s = t . So Zn(c) = { ( T , s , s ) E

Noting that imdn+l = {(O,t , t) E C,} for even n, we have H,(c ) = Z for

121

When n is odd, ( T , s , ~ ) E & ( c ) if and only if s + t = 0. So Zn(c) is isomorphic to Z $Z, and noting imd,+l = { ( ~ , 0 , 0 ) E C,} M Z, we have H n ( c ) = z .

Therefore, H,(c ) M Z for any n.

2305

Construct a CW complex which has the following 2-homology groups:

( Coz~mbia) Solution.

Denote by X1 the "figure eight space" as shown below,

Let X2 be the space obtained by attaching two 2-cells to X I , one by the map fi : S' + 20 and the other by the map f2 : S' + a such that f2(z) = z2. Therefore, the image of one 2-cell under f1 is homeomorphic to S2 , the 2- sphere. It is well-known that H1(X1) = Z$Z, which has two generators a and b, consider the following diagram

0 ---t HZ(X2) 5 H2(X2,X1) 3

a:

H'(X1) % H1(X2) - HI(X2, X') T f i * T f i l s ' +

0 --+ H2(D2,S1) + Hl(S1) + 0

The square is commutative and the level rows are exact. Since

H ~ ( x ~ , x') = imfl, CB imf2, M z CBZ,

122

imfllsi. = 0 and imf2Isi. = 2!J, we may see that ima, M 22 and kera, zY. Since H1(X2,X1) = 0, we have H1(X2) M Z $ Z 2 . It is also easy to see that

Now let X be the space obtained by attaching two 3-cells to X2, one by the map g1 : S 2 --+ 20 and the other by a map g2 : S2 -+ S 2 such that degg2 = 3. Then in a similar way as above, we may conclude that the space X satisfies the requirements in the problem.

H z ( X 2 ) =z.

2306

Compute H,(S" V S" V ... V S"), n 2 0, for all p . ( Columbia)

Solution. Denote by S"(q) the space S" V S" V ... V S"

J. q times

When n = 0, So(q) has

q + 1 points. Then H p ( S o ( q ) ) is a free abelian group of rank q for p equal to 0 and is zero otherwise. In fact

S"(q + 1) = S"(q) v S".

Let a E S n ( q ) and b E S". It is easy to see that U = Sn(q + 1) - { u } has the homotopy type of Sn(q) and V = Sn(q + 1) - { b } also has the homotopy type of S"(q). It is also clear that Sn(q + 1) = U U V and U n V has the homotopy type of S n ( q - 1). Thus, when n > 0, by induction on q and the Mayer-Vietoris sequence of the pair ( U , V ) , we may prove that gp(Sn(q)) is a free abelian group of rank q for p equal to n and is zero otherwise, where fip is the reduced homology group.

2307

Build a CW complex X by adding two 2-cells to S1, one by the map z --+ z4

(Indiana) and the other by the map z --+ z6. What is the homology of this space?

123

Solution.

Fig.2.14

X is a %dimensional CW complex with %skeleton K 2 = X, l-skeleton K1 = S1 and O-skeleton KO = { p } . K 2 is obtained from K1 by attaching two 2-cells via the maps f and g as indicated in Fig.2.14. K 1 is obtained from KO by attaching one l-cell. Let K" = K 2 for n 2 3. Then we have a chain complex K = {Cn(K),dn}, where Cn(K) = Hn(Kn,KnA1) and d, :

C n ( K ) -+ Cn-l(K) is defined to be the composition of homomorphisms,

where d, is the boundary operator of the pair (Kn,Kn-l) and j,-l is the homomorphism induced by the inclusion map. It is well-known that H,(X) M H n ( K ) . It is obvious that H n ( X ) = H,(K) = 0 for n >_ 3. To compute H2(h) , consider the following diagram:

HZ(K1) --t HZ(K2) 2 H2(K2, K1) 2 HI(K1) 2 H I ( K 1 , KO) T f* T (f IS^)*

H2(E2,S1) 3 Hl(S1)

The square is commutative and it is well known that f, is a monomorphism and 8; is an isomorphism. Noting flsi is the map z -+ z4, we see that the image of (flSl)* is 4Z. In the same way we see that the image of (glsi), is 6Z. Since

H ~ ( K ' , K ~ ) M imf, CB img,,

it follows that imd, is isomorphic to the subgroup 22 of H I ( @ ) , and that ker 3, is isomorphic to Z @&. Thus, since j 2 is injective, we see that

H 2 ( K 2 ) M imj2 M ker 8, M Z @&.

It is clear that j1 is an isomorphism. Therefore, imd2 M Z. Noting that ker d l % H1(K1, KO) M 2, we have H1(K) M 2/22 M &. Thus we conclude that

( zg3z2, i = 2 , i = 1, i = 0, Hi(X) =

otherwise. I 01

124

2308

Let X = {(z,y,z) E R3 I zyz = 0).

a) Compute H , ( X , X - { 0)) . b) Prove that any homeomorphism h : X + X must leave the origin fixed.

(Indiana) Solution.

a) Let U = D3 n X , where

D3 = {(z, y, z ) E R3 I z2 + y2 + z2 < 2).

It is easy to see that U is contractible. Thus

H , ( X , X - (0)) M H,(U, u - (0)) = H,-I(U - (0))

for any q. Let S = S2 n X , where S 2 is the unit 2-sphere. Then S is a deformation retract of U - (0). So H,(U - (0)) = g*(S) . To compute Z*(S), let

Wl = s - ( ( O , O , I), ( O , O , -1))

W2 = s - ((-11 0,0) , (1 ,0 ,0)) . and

Applying the Mayer-Vietoris sequence of the pair (W1, Wz), we see that

b) Let z be any point of X different from the origin. Similarly we obtain

Z @Z @Z, if z is in a coordinate axis, otherwise. H z ( X , X - { X I ) =

Let h : X + X be any homeomorphism. Since the local homology groups are invariant under homeomorphism,

H 2 ( X , x - (0)) M H 2 ( X , x - ( f (0))) .

From the above results, it follows that f (0) = 0.

125

2309

a) Write down the Mayer-Vietoris sequence in reduced homology which relates the spaces S1 x R1, S1 x R1 - { p } , and a disk D in S1 x R1 about the point p. (S1 = 1 - sphere)

b) Use a) to calculate the homology of S1 x R1 - { p } . ( Indiana)

Solution. a) The Mayer-Vietoris sequence is

-.. + Hq(D - { p } ) 3 Hq(S1 x R1 - { p } ) @ @,(D)

f gq(S1 X R1) -+ Hg-l (D - { p } ) -+ *. . . A -

b) Since S1 x R1 and D - { p } have the same homotopy type with S1, it follows that

Hg(S1 x R1) x H,(D - { p } ) x Hq(S1). Since D is contractible, the nontrivial part of the Mayer-Vietoris sequence is

0 + g1(D- { p } ) 3 x R1 - { p } ) f gl(S1 x R1) + 0, RZ %I

which is split exact. Thus we have

iYi(S1 x R1 - { p } ) = otherwise.

2310

Let the real projective plane RP2 be embedded in the standard way in the real projective 5-space RP5. Compute Hq ( R P 5 / R P 2 ) , where R P 5 / R P 2 is the space obtained from RP5 by identifying R P 2 to a point.

( Indiana) Solution.

Since RP5 is compact Hausdorff and R P 2 is a strong deformation retract of a compact neighborhood of R P 2 in R P 5 , we see that

H, ( R P 5 / R P 2 ) x Hq(RP5 , R P 2 ) for any q.

126

It is well-known that

z, Q = 0, { 0, otherwise, { 0, otherwise.

z, q = 0 , 5 , H,(RP5) = &, q = 1 ,3 , and H,(RP2) = &, q = 1,

Thus, from the exact sequence of the pair (RP5 , RP2) , i t follows that

a, 4 = 0 ,5 ,

0, otherwise. H , ( R P 5 / R P 2 ) =

2311

Let Y = {(z1,z2) E R2 I 2 1 2 2 = 0 and 2 2 2 0).

Prove that Y x R is not homeomorphic to R2. (Indiana)

Solution. Let

Y x R = { ( 2 1 , 2 2 , 2 3 ) E R 3 1 2 1 2 2 = 0 , 2 2 2 0 ) .

Then the point 0 = (0, 0,O) E Y x R. To compute the local homology groups

H i ( Y x R , Y x R - { 0 } ) , i = 0 , 1 , 2 , * . * ,

we take an open neighborhood of the point 0, U = D3 n (Y x R ) , where D3 is an open ball centered a t 0. Therefore

- Hi(Y x R , Y x R - (0)) M Hi(U, U - (0)) M H i - l ( U - { 0 } ) ,

because U is contractible. Let X be the space as shown in the following figure

127

It is easy to see that X is a deformation retract of U - (0) and that H I ( X ) X Z @Z. Thus

H z ( Y x R,Y x R - (0)) EZCBZ.

It is well-known that for any point p E R2 the local homology group

H2(R2 , R2 - ( p } ) GZ.

Since the local homology groups are isomorphic under homeomorphism, it follows that Y x R is not homeomorphic to R2.

2312

Let A be a nonempty subset of X and X u C A be the union of X with the cone of A; that is, X U C A is obtained from the subset X x (0) U A x [0, 11 of X x [0, 11 by identifying A x (1) to a point.

Prove that the reduced homology group g, (X U C A ) is isomorphic to H,(X, A ) , for every n.

Solution. Let Y = X x (0) U A x [0,1] and B = A x (1). Then X U C A = Y/B.

Let ?r : Y -+ Y / B be the identification map and yo denote the point ?r(B) in Y / B . Let U = A x [$,1] c Y . Then yo is a strong deformation retract of ?r(U). Thus, in the exact sequence of the triple (Y /B , ?r(U), yo),

(Indiana)

* . * + Hn(?r(U),YO) + H * ( Y / B , y o )

+ H,(Y/B,?r(U)) -+ H,-l(?r(U),YO) -+ * * *

it follows that H,(?r(U) , yo) = 0. Hence, the inclusion map of pairs induces an isomorphism

H*(Y /B ,YO) = H*(Y/B,?r(U)).

H * ( Y , U ) M H * ( Y - v, u - V ) .

Let V = A x [i, 11. By the excision property, we have an isomorphism

Since B is a strong deformation retract of U, it follows from the exact sequence that

H*(Y , B ) = H*(Y , U ) .

128

Thus, we have H * ( Y , B ) M H*(Y - v,u - V).

In a similar way the set T ( V ) may be excised from the pair ( Y / B , n-(V)) to give an isomorphism

Now the restriction of the map n- gives a homeomorphism of pairs

n- : (Y - v, u - V ) + ( Y / B - T(V) , T (V) - .(V)),

and so an isomorphism of their homology groups. All of these combine to give an isomorphism

H*(Y/B, yo) M H * ( Y , B) .

Since Y admits obviously a strong deformation retraction onto X x (0) which maps B onto A x { 0 } , we have

H,(Y, B ) M H , ( X , A) .

On the other hand, from the exact sequence of the pair ( Y / B , yo), it follows that

H*(Y/B,yo) x H*(Y/B) = G,(X UCA).

Therefore, H,(X U C A ) is isomorphic to H,(X , A ) for every n.

2313

For any topological space X let EX denote its (unreduced) suspension. (EX is the quotient space (X x [0,1]/ -, where - denotes the equivalence relation generated by requiring that (2, t ) - (y, s) if s = t = 0 or s = t = 1.) I f f : X + Y is a map, let Cf : C X -+ C Y be the map of suspensions induced by the map ( ~ , t ) -+ (f(z),t) of X x [0,1].

(i) Prove that if f : X -+ Y is a homotopy equivalence then so is Cf. (ii) Using only the Eilenberg-Steenrod axioms (Homotopy, Exactness, Ex-

cision, Dimension) for a homology theory prove that Hi (EX) is naturally isomorphic to Ei-l(X). (Here Hi denotes reduced singular homology.)

(Indiana)

1 29

Solution. (i) Denote by rIx : X x I + C X the quotient map, and by po, pl the

equivalence classes of (z, 0) and (z, l), z E X, respectively. Let f : X + Y be a continuous map. By the definition we see that

Cf = rIy 0 (f x id) 0 I IXl ,

where id : I + I is the identity map. Suppose that g : X + Y is another continuous map which is homotopic to f by a homotopy G : X x I + Y such that G(z,O) = f(z) and G ( z , 1) = g(z) for any z E X . Then define c : (X x I) x I -+ CY by

Let H : EX x I --t CY be a map defined by H ( Z , s) = E(II&'(%), s) for any (2, s) E EX x I , where, if Z = po or p l , lI,'(Z) means any point (z,O) or (z, 1). It is easy to check that H is well-defined and continuous. It is also easy to see that H ( Z , 0) = Cf(Z) and H ( E , 1) = Eg(Z). Therefore Cf is homotopic to Cg, and consequently it follows that if f : X + Y is a homotopy equivalence then so is Cf.

(ii) Let 1 3

2 3

u = {rIx(z,t) 1 t > -}

v = {rIx(z,t) 1 t < -}. and

It is obvious that U and V are both open sets of EX and that U and V are both contractible. Hence r?.i(U) = g,(V) = 0 for all i. It follows from the homology sequence of the pair (EX, U ) that fii(CX) M Hi(EX, U ) . Let W = {IIx(z, t ) 1 t 2 2/3}. Then c U . By the excision property we have

It is easy to see that U - W has the same homotopy type of X . On the other hand, from the homolopy sequence of the pair (V, U - W ) we see that

Hi(V, u - W ) M E&'(U - W).

Thus we conclude that k i ( C X ) k*-l(X).

130

2314

Consider the following commutative diagram of abelian groups in which the row and column are exact sequences.

Suppose that y and Po are surjective. Prove that a3 is injective. (Indiana)

Solution. Let a3 E A3 such that cyg(a3) = 0. Then from the exactness there is an

a2 E A2 such that az(u2) = a3. Since y is surjective, there is an a1 E Al such that y(a1) = /32(a2). Hence by the commutativity we have

so al(a1) - a2 E kerP2, and there is a bl E B1 such that P l ( b 1 ) = a(a1) - a2. Since PO is surjective, there is a bo E Bo such that Po(b0) = bl. Once again by the exactness we have 0 = PlPo(b0) = Pl(b l ) , which means that al (a1) - a2 = 0. Hence a3 = ~ 2 ( a 2 ) = a2a l (a l ) = 0. It means that kerag = 0, i.e., a3 is injective.

2315

Suppose that a topological space X is expressed as the union U U V of two open path-connected subspaces such that U n V is path connected, H 1 ( U ) = 0, and the inclusion UnV + V induces a surjective homomorphism H1(UnV) +

H1(V) . (Here H denotes singular homology.) a) Prove that H 1 ( X ) = 0.

131

b) Give an example to show that the analogous assertion becomes false in general if H1 is replaced everywhere by Hz.

Solution. (Indiana)

a) Consider the Mayer-Vietoris sequence of the pair ( U , V)

A - . . . + H ~ ( x ) 3 H ~ ( U n v) % H ~ ( u ) CB H~(v) 5 H ~ ( x ) + H ~ ( U n v). Since U n V is path-connected, Eo(U n V) = 0. It means H 1 ( X ) = im$*. By the hypothesis, d* is surjective so that

Therefore im& = 0, i.e., H 1 ( X ) = 0. b) Let X = S 2 , the unit 2-sphere7

1 u = {(zclg,z) E s2 I z > - 5 )

and 1 2 v = {(z, y, z ) E s2 I z < -}.

Then X = U U V . It is obvious that U , V and U n V are path-connected and that H z ( U ) = H2(V) = 0. So the homomorphism H z ( U n V) + H z ( V ) induced by the inclusion map is surjective. But H 2 ( X ) M Z.

2316

Let D" = {z E R" 1121 5 1)

denote the standard unit ball in Euclidean space Rn, and

denote the standard (n- 1)-sphere. Suppose that f : Dn --t D" is a continuous map such that the restriction of f to S"-l is a homeomorphism from S"-l to s n - 1 . Prove that f is surjective.

Solution. We use the reduction to absurdity. Suppose that there is a point 20 E Dn

such that zo @ f(D"). By the assumption, we see that 20 E D" - 9 - l .

(Indiana)

132

Therefore Sn-' is a deformation retract of D" - (20). Let T : D" - ( 2 0 ) -+

Sn-' be a retraction. Then g = r o f is a continuous map from Dn to S"-l and gIs.-1 : 9 - l -+ S"-' is just the restriction of f to Sn-', which is a homeomorphism. Therefore g o i : Sn-' -+ Sn-' is a homeomorphism. Hence

Let i : S"-l -+ Dn be the inclusion map.

(g 0 i)* : Hn-l(S"-') -+ H,-'(S"-')

is an isomorphism. But ( g o i)* = g* o i, and i, : Hn-1(Sn-') --t Hn- l (Dn) is obviously a trivial homomorphism because of Hn-l(Dn) = 0, and consequently (g o i)* is trivial. This is a contradiction.

2317

Let X be a connected CW complex with two 0-cells, three 1-cells, three 2-cells, and no higher-dimensional cells. Assume H1(X) x Z @Z/3. Compute the Euler characteristic of X and determine (with proof) all possibilities for H2(X).

(Indiana) Solution.

The Euler characteristic of X , X ( X ) = 2 - 3 + 3 = 2. On the other hand,

X(X) = rankHo(X) - rankHl(X) + rankH,(X).

Therefore rankHz(X) = 2. Let X k denote the k-skeleton of X . By the assumption, we see that H 3 ( X 3 , X 2 ) = 0, and H 2 ( X 2 , X 1 ) =Z$Z@Z. Therefore

H 2 ( X ) M ker{d, : H2(X2, X ' ) --t H1(X1 , XO)},

where d, is the composition of homomorphisms

H2(X2 , X ' ) 3 H1(X1) 2; H1(X1,XO).

Thus because H 2 ( X 3 , X 1 ) is free abelian, H 2 ( X ) is also a free abelian group of rank 2, i.e., H2(X) ZZ $Z.

133

2318

Let X be a CW complex with exactly one k+l-cell. Prove that if Hk(X;Z) is a nontrivial finite group then Hk+l(X;Z) = 0.

Solution.

position of homomorphisms of pairs

(Indiana)

Denote by Xk the Ic-skeleton of the CW complex X , and by dk the com-

H k ( X k , x k - 1 ) 3 Hk-l(Xk-l) 5 H k T l ( X k - l , x k - 2 ) .

It is well-known that i Y k ( X ; Z ) M ker dk/imd"'

and Hk+l(X;Z) M ker dk+1/imdk+2.

By the hypothesis, H ' + l ( X k + l , X k ) zz.

We choose a generator a in Hk+l(Xk+l ,Xk). We claim that &+'(a) # 0. Otherwise we would have imd'+l = 0. Therefore

H k ( X ; Z ) M ker dk c H k ( X k , X"-').

But it is well-known that Hk(Xk,Xk-l) is a free abelian group, and, consequently, kerdk is trivial or free abelian. It means that H k ( X ; Z ) is not a nontrivial finite group and contradicts the hypothesis. Thus we have d'+'(a) # 0. It follows that kerd"' = 0 and, therefore, Hk+l(X;Z) = 0.

2319

Let

and

134

Let X = A u B. Compute the relative homology groups H i ( X , X - (0, 0, 0,O)) for all a.

Solution.

U = 6 n X is an open neighborhood of ( O , O , O , 0) in X . property we have

(Indiana)

Let 6 be the unit open ball centered at the point ( O , O , O , O ) in R4. Then By the excision

&(X, x - ( O , O , 0,O)) M Hi(U, u - ( O , O , 0,O))

for all i. It is obvious that U is contractible and that U - (0, 0, 0,O) has a deformation retract C U D, where

and D = { ( ~ 1 , 2 2 , 2 3 , 2 4 ) E R4 I 24 = 0, 2: + 2: + 2: = 1).

(See Fig.2.15)

Fig. 2.15

Let Wi = C U D - {pl,p2) and W2 = C U D - {p3,p4}. (See Fig.2.15.) Then W1 and W2 are open subsets of C U D and W1 u W2 = C u D. It is clear that both C and D are homeomorphic to the unit sphere S2 and that D and C are deformation retracts of W1 and Wz respectively. I t is also clear that S1 is an deformation retract of W1 n Wz. Therefore, applying the Mayer-Vietoris sequence to the pair (W1, W2), we see that

2 = 0, other wise.

From the homology sequence of the pair ( U , U - (0, O , O , 0)), we see that

Hi+l(U, u - ( O , O , 0,O)) = Hi(U - ( O , O , 0,O))

135

for all i 2 0. Hence we conclude that

H j ( X , X - ( O , O , O , 0)) = i = o , 1 otherwise.

2320

Let X be a path connected and locally path connected space and let x E X . Let Y = S1 x S1 x S2. Show that if al(X,z) is finite, then any continuous map f : X -+ Y induces the trivial map f* : H i ( X , Z ) + H i ( Y , I ) for all i different from 0 and 2.

(Indiana) Solution.

It is easy to see that R2 x S2 is the universal covering space of Y . By the Kunneth theorem, we have

It is clear that

Rl(S1 x s1 x S1) = “l(T2) @ “l(S2) = a $a. Since ?rl(X, x) is finite, the homomorphism f* : R ~ ( X , z) --+ R ~ ( Y , f(z)) must be trivial. Therefore, we may lift f to a map 7 : X -+ R2 x S 2 such that a o f = f, where R : R2 x S 2 -+ Y is the covering map. Hence we have

- -

f* = R, 0 f* : H l ( X , Z ) -+ H i ( Y , Z )

for any i. But it is obvious that

: H i ( X , Z ) -+ H i ( R 2 x S 2 , Z )

is trivial for i different from 0 and 2, so is

136

2321

Call a commutative diagram of abelian groups

“exact” if the sequence of groups and homomorphisms

o + A (9) B c~ c ’5‘ D - o is exact.

It is an interesting fact that if

A Z B

C + D P 1 17

6

and B S E

D - F 71 1 E

6’

are exact, then so is A Cl’a - + E

C - t F P I 1 E

6’6

Prove exactness of

(Indiana) Solution.

By the assumptions, we have 6P = 7a and &a’ = 6’7. Therefore, we have

(6’6 - &)(P,a’a) = 6’6P - &a’a = 6’ya - 6’ya = 0.

1 37

It follows that im(P,a’a) c ker(6’S - E ) . Let ( c , e) E ker(6’6 - E ) , i.e., S’6(c) - E(e) = 0. By the assumptions, we have im(a,p) = ker(y - 6) and im(a’,y) = ker(c - a’). Thus, noting that

( e , 6 ( c ) ) E ker(E - S’),

we see that there exists a b E B such that a’(b) = e and y(b) = 6 ( c ) . Hence we have (b , c ) E ker(y - 6). Therefore, there exists an a E A such that a(.) = b and p(a) = c, and, consequently,

( c , e ) = (0, a’a)(u) E im(0, a’a).

It means that ker(6’6 - E ) c im(P, a’a).

The exactness at C @ E is proved.

2322

Let X be a non-empty compact Hausdorff space and f : X --+ X be a continuous map. Prove that there exists a non-empty closed subset A of X such that f ( A ) = A . Give an example to show that compactness is essential for this assertion.

Solution. Let F1 = f ( X ) . Then F1 is a non-empty closed subset of X. We define

Fn+l = f(Fn) inductively for n E Z+. It is clear that { F n , n E Z+} is a sequence of non-empty closed subsets in X and that

(Indiana)

00

Since X is compact, we see that the subset A = n F, is a non-empty closed

subset of X . We claim that f ( A ) = A holds. We give an example to show that compactness is essential for this assertion.

Let X = (0,1] and f : X --+ X is defined by f(z) = i z for z E X. Suppose that a non-empty closed A of X satisfies f ( A ) = A. Then there would exist an 2 0 E A such that z 5 zo for any z E A. In fact z o = sup z. Since f ( A ) = A

and zo E A , there would exists an z1 E A such that f(z1) = 20, i.e., 20 = 521. Therefore zo 2 z1 = 220, which is a contradiction.

n = l

xEA 1

138

2323

Recall that H 3 ( S 3 ; Z ) x H3(RP3;Z) MZ.

(RP" is n-dimensional real projective space.) Prove that there is no function f : S3 + RP3 inducing an isomorphism on the third homology.

Solution. It is well-known that S3 is the 2-fold universal covering space of RP3.

Let 7~ : S3 + RP3 denote the universal covering map. Then it is clear that the degree of 7r is equal to 2. Let f be a function from S3 to RP3. Since 7r1(S3) = { 0 } , there is a lifting of f, f : S3 + S3 such that 7 r f = f . Denote by a and b the generators of H 3 ( S 3 ) and H3(RP3) respectively. Then we have

f*(a) = 7r*(f*(a)) = degf . r , ( a ) = 2-deg f . b ,

because deg f E Z, it is obvious that f* is not an isomorphism.

(Indiana)

- I

2324

Let x = ((2, Y, .> I ZY = 01.

(a) Compute H1(X - ( O , O , 0 ) ) . (b) Using part a , show that X is not homeomorphic to R2. (c) Prove or disprove: X is homotopy equivalent to R2.

(Indiana) Solution.

(a) In fact, X = 7 ~ 1 U 7r2 where 7r1 is the plane y = 0 and 7r2 is the plane z = 0. Denote by A and B the unit circle in 7rl and a2 respectively. Then it is easy to see that A U B is a deformation retract of X - ( O , O , 0). Take U = A U B - (0, 0 , l ) and V = A U B - (0, 0, -1). Therefore, U and V are both contractible, U U V = A U B and U n V has four path components which are all contractible. Applying the Mayer-Vietoris sequence to the pair ( U , V), we see that

H1(X - ( O , O , O ) ) x H1(A U B ) =Z $Z $Z.

(b) Suppose that there exists a homeomorphism f : X + R2. Then f lx- (o ,o ,o) : X - ( O , O , 0 ) + R2 - f ( O , O , 0) is also a homeomorphism, which

139

wouid induce an isomorphism from H1(X - ( O , O , 0)) to H1(R2 - f ( O , O , 0)). But it is clear that

It is a contradiction. (c) Let F : X x I + X be a map defined as follows.

Then F is a deformation retraction of X onto the plane 7r2.

homotopy equivalent to R2. Hence X is

2325

Let (B" , Sn-l) be the standard ball and sphere pair in R", n > 1. Suppose that f : ( B " , S n - l ) -+ ( X , A ) is a continuous map and flsn- l : S"-l -+ A is a homeomorphism. Show that if & ( X ) = 0 then H , ( X , A ) = Z .

Solution. (Indiana)

Consider the following diagram

a. i* H , ( X ) -1?;H,(X,A) + H , - i ( A ) ---t H , - i ( X ) t f* t fl* t f*

a: i: Hn(B",S"-1) + H " - l ( S n - l ) + Hn- l (B")

in which the level rows are exact and the squares are commutative. By the assumption, f1, is an isomorphism, and therefore

It is well-known that 8: is an isomorphism and Hn- l (Bn) = 0. Thus we have

keri, = H , - l ( A ) = Z .

Since H , ( X ) = 0, d, is a monomorphism. Hence we have

H , ( X , A ) M imd, M ker i, = Z.

140

2326

(a) Describe a CW structure on S2 x S5 and use it t o compute the homology

(b) Compute the homology of S2 x S5 with 2 points removed. of s2 x s5.

( I n d i a n a ) Solution.

(a) Let xo E S2, x1 E S5 . Then S2 is obtained by attaching a 2-cell t o 20, and S5 is obtained by attaching a 5-cell to XI. Denote by S2 V S5 the one point union of S2 and S5, which can be considered as the space obtained by attaching a 2-cell and a 5-cell to the point (20, 21) E S2 x S5. It is easy to see that S2 x S5 is homeomorphic to the space obtained by attaching a 7-cell to S2 V S5. Therefore the CW structure on S2 x S5 has a 0-cell, a 2-cell, a 5-cell and a 7-cell. Hence it is obvious that H i ( S 2 x S5) is an infinite cyclic group for i equal to 0, 2, 5, and 7, and is zero otherwise.

(b) Let X = S 2 x S 5 - { p l , p 2 } , U = S2xS5-{p1}andV = S 2 x S 5 - { p 2 } . Then U , V are both open sets of S2 x S5 and UnV = X, and UUV = S2 x S5. It is easy to see that S2VS5 is a deformation retract of both U and V . Applying the Mayer-Vietoris sequence to the pair (U, V) and using the fact that

we conclude that

i = 0,2,5,6, ez ez ez, i = 1,

otherwise.

2327

Let A be a subspace of S 2 x S2 homeomorphic to the 2-sphere S2. State what the homology groups H , ( S 2 x S2) are (no proof is required). What can you say about the possibilities for the relative homology groups H,(S2 x S2, A)?

Solution. ( I n d i a n a )

By the Kiinneth formula we conclude that

otherwise.

141

From the homology sequence of the pair ( S 2 x S 2 , A ) , using the above result and the fact that H;(A) M Hi(S2),;we see that

H ; ( S 2 x S 2 , A ) M H i ( S 2 x S 2 )

for i 2 4 and that the nontrivial parts of the sequence are

0 4 H3(S2 x S 2 , A ) 4 H 2 ( A ) 5 H 2 ( S 2 x S 2 ) 5 H 2 ( S 2 x S 2 , A ) -+ 0

and

a. - 0 + H1(S2 x S 2 , A ) -+ Ho(A) -+ Ho(S2 x S 2 ) -+ H o ( S 2 x S 2 , A ) -+ 0.

Since &(A) = 0, H,(S2 x S2,A) = 0 for i equal to 0 and 1. By the exactness, we have H3(S2 x S 2 , A ) M keri, and H 2 ( S 2 x S 2 , A ) M Z @Z/irni,, where i, is the homomorphism induced by the inclusion map i : A - S2 x S 2 .

2328

Compute the homology groups of the space X obtained as the union of the 2-sphere S2 and the x-axis in R3.

Solution. Let

(Indiana)

X’ = X - {the open interval (-1,l) in the x-axis}.

Then the space X may be viewed as a space obtained from X* by attaching a 1-cell. It is clear that Hi(X, X”) is infinite cyclic for i equal to 1 and is zero otherwise. It is easy to see that X’ has the homotopy type of 5’. Thus from the homology sequence of the pair ( X , X”) we conclude that Hi(X) is infinite cyclic for i equal to 0,1,2 and is zero otherwise.

2329

Define the “unreduced suspension” CX of a space X to be the quotient space of I x X obtained by identifying (0) x X to one point and (1) x X to one point. (This is the union of two “cones” on X.) Show that there is a natural isomorphism ax : H,(X) 4 Hi+l(CX) , for all i 2 0.

(Indiana)

1 42

Solution. Let (0)’x X and (1) x X be identified to the points 20 and 2 1 respectively.

Take U = EX - ( 2 0 ) and V = C X - (21). Then U and V are open sets of E X , and U u V = EX. Consider the Mayer-Vietories sequence of the pair (U, V ) :

-.. -, H ; + ~ ( U ~ V ) 3 H i + l ( u > ~ H i + l ( ~ > s H ~ + ~ ( E x ) -+ Hi(unv) -+ .... 9 - A - -

It is clear that U and V are contractible and that U n V has the homotopy type of X . Thus the homomorphism A : k i + l ( C X ) + k i ( U n V ) is an isomorphism and we may obtain an isomorphism A* : Hi+l (CX) --t &,(X) . The inverse of A*, CTX : & i ( X ) -+ Hi+l (CX) is just what we are looking for. The naturality of ax can be derived from the naturality of the Mayer-Vietories sequence.

-

2330

Denote by X the union of the torus S1 x S1 with the disc D 2 , where D2 is attached to T2 by identifying dD2 with a meridian curve S1 x (20) in the torus, where 20 E S1. (See below.)

2 0

Fig.2.16

(a) Calculate H,,(X) for all n 2 0. (b) Is T2 a retract of X ? Why or why not?

(Indiana) Solution.

(a) Consider the homology sequence of the pair ( X , T 2 ) . It is well-known that H ; ( X , T 2 ) = 0 for i # 2 and H 2 ( X , T 2 ) M f * (H2(D2 ,dD2) ) zZ, where f * is the homomorphism induced by the adjunction map f : D2 -+ X . Thus the nontrivial part of the sequence is as follows.

0 4 H 2 ( T 2 ) 5 H 2 ( X ) 1; H 2 ( X , T 2 ) a: -+

Qt f* t f l*

H1(T2) 2 H 1 ( X ) -+ 0

0 + H2(D2, a D 2 ) 4 H l ( d D 2 ) -+ 0 w

143

It is easy to see that ima: = irnfl. and ker& = ker f 1+ .

H 1 ( T 2 ) = Z @Z is generated by a and b. f1*(Hl(aD2)) = 2 @ (0). H z ( X ) M H2(T2) xZ. Hence we conclude that

We know that (See Fig.2.16) It is clear that

Thus we see that H 1 ( X ) x Z @Z/Z M Z and

(b) We claim that T2 is not a retract of X. Otherwise, there would exist a retraction map T : X .+ T2 such that ~ ( p = i d p . Let i : T2 + X be the inclusion map. Then r o i = i&z : T 2 -+ T2 , and, consequently, we have T* o i, = id : H1(T2) - H1(T2) . In other words, we have the following commutative diagram

because of H 1 ( T 2 ) = Z @I and H 1 ( X ) =Z. This is a contradiction.

2331

Given homomorphism h : A -+ B and g : C -+ B , the pull back of h via g is the group

g*(A) = { ( c , a ) E C x A 1 g(c) = h(a)} .

(a) Let g*(h) : g*(A) + C be the homomorphism obtained by restricting the projection onto the first factor C x A -i C to g*(A) . Prove that the kernel of g*(A) is isomorphic to the kernel of h.

9* ( A ) A 1 g*(h) l h

- + B 9

c

(b) Let A = Z / U , B = 2/Z, and let h : A -+ B be the surjection defined by h(1) = 1. Let C = 2/82 and let g : C -+ B be the surjection defined by g(1) = 1. Identify the group g*(A) , with explanation.

(Indiana)

144

Solution.

C = l,, the identity of C. By the definition of g*(A), we see that (a) Suppose that (c ,u ) E kerg*(h). Then g*(h)(c,a) = l,, and, therefore,

kerg*(h) = {(l,,a) I a E kerh}.

Let F : kerg*(h) -+ ker h be defined by F(l,, a ) = a. It is easy to see that F is an isomorphism.

(b) In this case, we have h-'(O) = {0,2}, h-'(l) = {1,3}, g-'(O) = {0,2,4,6} and g-'(l) = {1,3,5,7}. Thus it is clear that the group g*(A) consists of the following 16 elements: ( O , O ) , (0,2), (2,0), (2,2), (4,0), (4,2), (6,0), (6,2), (1,1), ( L 3 L (3,1>, (3,3), (5, I), (5,3), (7, I>, (7,3).

2332

Recall that if C is a homeomorphic copy of the circle in S3, then Hi(S3 -C) is infinite cyclic for i equal to 0 or 1 and is zero otherwise. Assuming this fact compute

(a) The homology of R3 - C, when C is a homeomorphic copy of the circle in R3.

(b) The homology of Y = R3 - X , where X c R3 is a homeomorphic copy of the "figure-eight space" (i.e., the one-point union of two circles.)

Solution. (a) Denote S3 = R3 U {a}. Let A = S3 - C, B = S3 - {m}. Then A and

B are open subsets in S3, and A U B = S3 and A n B = R3 - C. We have the following Mayer-Vietoris sequence.

(Indiana)

- ... + H,+'(S3) 2 Hi(R3 - C ) % Hi(S3 - C ) @ Hi(S3 - {co})

A - % Hi(s3) -+ H ~ - ~ ( R ~ - C ) + -. .

Noting that Ga(S3 - {a}) = G@3) = 0

for any i and that

and

z, i = 3, H i p 3 ) = { 0, otherwise,

Hi(S3 - C ) =

145

we can see that i = 0,1,2,

- = { :: otherwise.

Fig.2.17

Represent X as shown in Fig.2.17. Let U = X - {PI} and V = X - ( ~ 2 ) .

Then we see that U and V have homotopy type of S 1 and that U n V is contractible. Since U u V = X , we have

Y = R~ - x = ( R ~ - u) n ( R ~ - v)

( R ~ - u) u ( R ~ - v) = R~ - (u nv). and

From the result of (a), we see that Zi(R3 - U ) and gi(R3 - V ) is infinite cyclic for i equal to 1 or 2 and is zero otherwise. It is easy to see that

iii(x3 - (v n v)) M iri(s2)>. Due to the above facts, the nontrivial part of the Mayer-Vietoris sequence

of the pair (R3 - U, R3 - V ) is

o --+ i i2(y) 2 i i 2 ( ~ 3 - U ) CB i i 2 ( ~ 3 - V ) 5 i i 2 ( ~ 3 - u n V > %Z$Z MZ

A - + H 1 ( Y ) 2 i i1(R3 - U ) @ i i1(R3 - V ) --+ 0.

%A7 @Z

We claim that the homomorphism is an epimorphism. Take a sufficiently large T > 0 such that the sphere S 2 ( ~ ) belongs to R3 - X . Then S 2 ( r ) is a deformation retract of R3 - U n V . Hence we can consider the generator of H 2 ( S 2 ( r ) ) , [c], as the generator of E2(R3 - U n V). Since the representative chain c of [c] can also be considered as a chain of (R3 - U ) n (R3 - V ) , by the definition of ++ we have [c] = ++( [c] , 0). The claim is proved, which means imA = 0 too. Thus we get H 1 ( Y ) =Z @Z and a split exact sequence

-

-

O+&(Y) 5ZCBZ5Z--+O.

146

It follows that HZ(Y) =Z. So we conclude that

i 2 3, - i = 2, H l ( Y ) = {:w Z7 i = l ,

I i = 0.

2333

It is known that if X c S3 is homeomorphic to S1 then H , ( S 3 - X ) M

H,(S1). Use this fact to compute the homology of S3-Y where Y is a subspace of S3 homeomorphic to the disjoint union of two copies of S1.

Solution.

morphic to S1. Therefore,

(Indiana)

Denote Y by Y = A U B, where A n B = 8 and both A and B are homeo-

S3 - Y = ( S 3 - A ) n ( S 3 - B) .

Noting S3 - A and S3 - B are open sets of S3 and

( S 3 - A ) U ( S 3 - B) = S 3 ,

in the Mayer-Vietoris sequence we have

A - * . . -, ii ,+l(~3) -, H , ( s ~ ) -, H , - ~ ( s ~ - Y) -, . .

H , ( s ~ - Y) 5 &(s3 - A ) @ &(s3 - B) *. - A - 4.

Using the fact that

and that

Z, 4 = 1, 0, 4 # 1,

Z,(S3 - A ) M Hq(S3 - B ) M H,(S1) =

i i , (S3) = { Z, 4 = 3, 0, q # 3 ,

we can easily see that

4 = 2 , - { 2*Z, 4'1, H,(S3 - Y) =

1 47

2334

(a) Sketch pictures of the universal covering of the one point union S1 v S2

(b) Compute the homology of the connected 2-fold covering space of S1vS2. and of the connected 2-fold covering (no proofs required).

(Indiana) Solution.

(a)

...

Fig.2.18

The universal covering of the one point union S1VS2 is shown as in Fig.2.18. The covering map T , restricted on each S2, is the identity map, and, restricted on R, is the exponential map: R + S1. The connected 2-fold covering of S1 v S2 is shown as follows.

Fig.2.19

The covering map T , restricted on each S2, is the identity map, and, re-

b) To compute the homology of the connected 2-fold covering space of stricted on S1, is the 2-fold covering map: z + z2 from S1 to S1.

S1 v S 2 , i.e., the homology of S2 V S1 V S 2 , we take

U = S2 V S1 V S2 - {the antipodal point of p l }

and V = S2 V S1 V S2 - {the antipodal point of p 2 }

(see the above figure). Then it is easy to see that S1 is a deformation retract of U n V , and that S1 V S2 is a deformation retract of U and V.

Thus, it is clear that

148

Z, 4 = 1121 c 0, 4 # 172.

and & ( U ) M &(V) x H,(S2) @ H,(S1) =

Therefore, the nontrivial part of the Mayer-Vietoris sequence is

A - o + H2(u) @ H2(v) 5 H2(s2 v s1 v s ~ ) + H ~ ( U n v) 5 & ( U ) @ &(V) 2 &(S2 v s1 v S 2 ) + 0.

It is easy to see that the inclusion maps k : unv + U and I : UnV + V induce injective homomorphisms k, : H1(U n V) -+ k 1 ( U ) and I , : H1(U n V) +

H1(V), respectively. Hence the homomorphism

- - -

4, : G1(u n v) + H1(u) @ H1(v)

is injective, i.e., ker 4% = 0. Thus imA = 0. It means that - - H z ( S 2 V S 1 V S 2 ) ~ k e r A z i m $ , z H 2 ( U ) @ H z ( V ) = Z @ Z .

It is clear that - - H ~ ( s ~ v s1 v s2) M H ~ ( u ) kl(V)/im#, MZ.

Hence, we have

H,(S2 v s1 v S 2 ) = 4 = 1, otherwise.

2335

Compute the homology of S1 x S1-point.

Solution.

Fig. 2.20

(Indiana)

It is easy to see that the space X = A u B is a deformation retract of the space H = S1 x S1-point, where A and B are each homeomorphic to S1 and

1 49

A n B = {zo} as shown in Fig.2.20 Choose points a E A and b E B such that u # z o and b # 20. Let U = X - { b } , and let V = X - { u } . It is clear that A and B are deformation retracts of U and V , respectively, and that U n V = X - {a ,b} is contractible. Applying Mayer-Vietoris sequence t o U and V, we have

n L 2 , Hn(S1 x S1 - point) = H , ( X ) = H,(A) @ H,(B) =

2336

Suppose the following diagram is commutative, the rows are exact, and 7, is an isomorphism for all n.

Construct an exact sequence

... -+ A, +A’ , @ B, -i BL + An-l -+ * * *

Write out the proof of exactness at Bh.

Solution. The following sequence is exact:

(Indiana)

where A is defined by A(a’, b ) = ;;(a’) - P n ( b )

for any (a’, b) E A; @ B,. We give the proof of exactness at Bh as follows. Let u E ker(6, 07;lojA). Then 7;’ o j L ( u ) E ker6,. Due to the exactness

a t C,, there exists a b E B, such that jn(b) = 7;’ o j ; (u ) , and consequently, 7, . j n ( b ) = jA(b) . From the commutativity, we have j ; o@,(b) = jk(u) . Thus Pn(b) - u E kerj; and there exists an a’ E A; such that i;(a’) = Pn(b) - u. It means that

u = P,(b) - i;(d) = A(-a’, -b).

150

So ker(6, o 7;’ o j ; ) C imA. Now we prove

imA c ker(6, o 7;’ ojk).

Suppose that u E imA, i.e., u = ih(a’) - P,(b) for some a’ E Ah and 6 E B,. Since j ; o ih = 0 and j ; o Pn = 7, o j , , it is easy to see that

6, 0 7,’ 0 j L ( U ) = -6, 0 j ,(b) = 0.

Thus the exactness a t B; is proved.

Suppose that X is a space and f : X + Y , g : X + 2 are two maps of X into contractible spaces Y and 2. Let M be the mapping cylinder o f f and g , that is, M is the identification space obtained from the disjoint union of Y , X x I and Z by identifying each (z,O) with f (z), and each (z, 1) with g(z). Prove that H,M Z H,-1X.

Solution. II I1 Let U = X x [0,3/4] = Y/ -, where X x [0,3/4] = Y denotes the disjoint

union of X x [0,3/4] and Y , the equivalence relation - is determined by (z, 0) - II f(z). In the same way, let V = X x (f,1] = Z/ -’, where the equivalence

relation -’ is determined by (z, 1) - g(z). Then we have M = U U V and U n V = X x (1/2,3/4). Noting U, V and U n V are open sets of M , by Mayer-Vietoris sequence, we have the following exact sequence:

- - (Indiana)

A - * . . + H,(u n V ) + H J U ) @ I?p(v) + H,(M) + H , - ~ ( u n v > -+ H q - 1 ( U ) @ Hq- l (V) + ’..

- - (1)

II Since X x (0) = Y / - is homeomorphic to Y and is a deformation retract of U , from the assumption that Y is contractible, we have H,(U) 2 H,(Y) = 0. In the same way, H,(V) E H,(Z) = 0.

It is obvious that U n V has the same homotopy type with X . So k , (U n V ) Z & ( X ) . Thus, from (l), k , ( M ) Z k q - l ( X ) .

- - - -

Part 111

Differential Geometry

153

SECTION 1 DIFFERENTIAL GEOMETRY OF CURVES

3101

Let a ( s ) be a closed plane curve. Define the diameter d, of a ( s ) to be

Now assume that the curvature k ( s ) 2 1 for all s. i) For any N EZ+ sketch an example of such an a(.) with d, > N . ii) Assume further that a is a simple closed curve. Prove that d, 5 2 (or

some other constant independent of a; 2 is the best possible such constant).

Solution. (Indiana)

i) The following is an example of such an a ( s ) with k ( s ) 2 1 and d, > N .

"t

Fig.3.1

ii) From the hypothesis that k ( s ) 2 1 for all s, we know that the simple closed curve a is an oval.

For every oval, by Blaschke, if we take the origin 0 as shown in the figure and denote by p ( 0 ) the distance from 0 to the tangent 1 at the point (z,y) of the oval, where the oval is counterclockwise orientated and 8 denotes the oriented angle from the z-axis to I, then the oval can be parameterized by 8 as follows

.(e) = p(e ) sin 6 + p'(0) cos 8, y(e) = -p(e) cos e + pye) sine.

154

p ( 8 ) is called the support function of the oval. From this we can conclude that, by direct computation, the relative curvature of the oval is k,.(8) = (p (8 ) +

(8) ) - 1.

Fig.3.2

Now we can prove a more general result of Blaschke: Let two ovals C and C1 in a plane be internally tangent a t a point 0.

Suppose that , a t every pair of points P and PI where C and C1 have the same tangent orientation, the curvatures of C and C1 satisfy the inequality k l (Pl ) 5 k(P) . Then the domain encircled by C1 must contain the domain encircled by C.

Fig.3.3

In fact, take the tangent point 0 of C and C1 as the origin, and their common tangent line as the z-axis. Let p ( 8 ) and p l ( 8 ) be the support functions of C and C1, respectively. From the above, we can say that the support function p ( 8 ) must be the solution to the following initial value problem of ODE

because p ’ ( 8 ) is exactly the distance from 0 to the normal line of C at point (.(a), y(8)). Hence, p ( 8 ) can be uniquely determined by k ( 8 ) = k, . (8) of C,

155

Analogously, for the curve C1, we also have the similar expression of p l ( 8 ) . Thus,

By the hypothesis, we know that

As for the sign of sin(8 - b) , firstly, if 0 5 8 5 x, then owing to 0 5 4 5 8, we have sin(8 - 4) 2 0. Therefore, p l ( 8 ) - p ( 8 ) 2 0. Secondly, if x 5 8 5 2x, we make the reflections of the oval C and C1 with respect t o their common tangent line a t 0 and reverse the orientation of the z-axis. Then we can get pl(8) - p ( 8 ) 2 0, where 8 and its corresponding original 8 satisfy 8 + 8 = 27~. Hence, we always have p l ( 8 ) 2 p ( 8 ) . Noticing that every oval is the envelope of all its tangent lines, we see that the domain encircled by C must be contained in the domain encircled by C1. The assertion of Blaschke is proved.

If we take a circle with radius 1 and centered at a ( s g ) + N ( s o ) as C1, and take a as C, then ii) follows immediately from the above assertion.

- I - -

3102

Let a be a regular C" curve in lR3 with nonvanishing curvature. Suppose the normal vector N ( t ) is proportional to the position vector; that is, N ( t ) = c( t )a( t ) for all t , where c is a smooth function. Determine all such curves.

Solution. For convenience, we assume that the parameter t is the arc length of the

curve a. Differentiating both sides of the equality N ( t ) = c(t)cr(t) with respect to t , we have, by the Frenet formula,

(Indiana)

Thus we can immediately obtain k ( t ) = constant, T ( t ) = 0 and c ( t ) = -k. Therefore, by the fundamental theorem of the theory of curves, we know that the curve a must be a circle (or a part of it).

156

3103

A surface S c lR3 is called triply ruled if a t every p E S we can find three open line segments L1, Lz, L3 lying in S such that L1 n L2 n L3 = {p} . Determine all triply ruled surfaces.

Solution. If S is a triply ruled surface, then, by the hypothesis, there are three differ-

ent asymptotic directions at every point of s. Observe that every asymptotic direction ( d u , dv) satisfies

( Indiana)

L(u, w)du2 + 2 M ( u , v)dudv + N ( u , v)dv2 = 0,

where L(u, v), M ( u , v), N(u, v ) are the coefficients of the second fundamental form of S at point p ( u , v ) . Noticing that the above equation is of 2nd order with respect to du : dw and it has two roots, we obtain L(u, w) = M ( u , v) = N ( u , v) = 0 for all (u , v). In other words, every point of S is a planar point. Therefore, S must be a plane (or a part of it).

3104

Let y : (a ,b) 4 R3 be smooth with [y’l = 1 and curvature k and torsion 7, both nonvanishing. Denote the Frenet frame by { T , N , B } . Assume there exists a unit vector a E R3 with

T . a = constant = cos a.

a) Show that a circular helix is an example of such a curve. b) Show that N . a = 0. c) Show that k / r = constant = k t a n a .


a) Let a circular helix be parameterized as follows

y(s) = (Tcosws,Tsinws, hws),

157

where T , h, w = ( r2 + h 2 ) - i are all constants. Then it is easy to verify that s is the arc length parameter. Hence

T ( s ) = (-rwsinws,rwcosws,hw).

If we take a = ( O , O , l), then T ( s ) . a = hw = constant. b) Differentiating T e a = constant with respect t o the arc length parameter

s, we obtain k ( s ) N ( s ) .a = 0, from which follows N ( s ) . a = 0 for all s E (a , b ) . c) By the property of b), we may assume that the constant vector

a = c o s a . T ( s ) f s i n a . B ( s ) .

Differentiating the above equality with respect to s, we have

0 = (cos a 3 k( s) f sin a . T( s ) )N (s) .

Thus, for all s E ( a , b ) , k ( s ) / ~ ( s ) = & t a n a .

3105

Show that if y is a geodesic on the cone t = d m , (z) y) E R2\{0, 0)) then y intersects itself a t most a finite number of times.

Solution. If we cut the cone along a generator 1 and develop it into a plane, then the

cone becomes an infinite sector without the vertex, and the geodesic y becomes a straight line on the developed infinite sector. Noticing that the central angle of the sector is f i r , an obtuse angle, and the image of y on the sector must be one of the following three cases:

(Indiana)

a generator, a straight line never intersecting the generator 1, two rays which start from the generator I ,

then we can conclude that y never intersects itself.

3106

Let y : (a ,b) 4 R3 be a C" curve parameterized by arc length, with curvature and torsion k ( s ) and ~ ( s ) . Assume k ( s ) # 0, ~ ( s ) # 0 for all s E (a ,b) , and let T and N denote the unit tangent and normal vectors to

158

y. The curve y is called a Bertrand curve if there exists a regular curve 7 : (a ,b) + IR3 such that for each s E (a ,b ) the normal lines of y and 7 a t s

are equal. In this case, 7 is called the Bertrand mate of y and we can write y(s) = y(s) + T N ( s ) for some T = ~ ( s ) E IR. (Note that s might not be an arc length parameter for 7.)

(a) Prove that r is constant. (b) Prove that if y is a Bertrand curve (with r as above), then there exists

(c) Prove that if 7 has more than one Bertrand mate, then y is a circular a constant C such that r k ( s ) + CT(S) = 1 for all s E (a , b) .

helix.

Solution. (Indiana)

(a) From y(s) = y(s) + r ( s ) N ( s ) it follows by differentiation that

Taking inner products a t both sides with N ( s ) = fN(s), we have that r’(s) = 0, namely r = const.

(b) From (a), now we have

- ds ds ds ds

T ( s ) = -(1- r k ( s ) ) T ( s ) - - T T ( S ) B ( S ) .

- If we denote T(s ) = a ( s ) T ( s ) + b ( s ) B ( s ) ,

then by differentiating with respect to s, we obtain

- dZ ds

k ( s ) N ( s ) - = a’(s)T(s) + ( a ( s ) k ( s ) + b(s) . r ( s ) )N(s) + b’(s)B(s).

Hence, from N ( s ) = fN(s) we know that a’(s) = b’(s) = 0, namely

ds ds -(1 - r k ( s ) ) = const, dz dz - r ~ ( s ) = const.

Therefore, there exists a constant C such that r k ( s ) + CT(S) = 1 for all s E

(c) Suppose that y1 and 7 2 are the Bertrand mates of y. Then, by (b), ( a , b) .

there exist constants T I , rg, C1, C2 such that for all s E (a , b) ,

r1k(s) + ClT(9) = 1, r2k( s ) + C27(5) = 1.

159

Because the non-zero constants T I , 1-2 are not equal, the above system of linear algebraic equations has solution k ( s ) = const, ~ ( s ) = const. Hence y must be a circular helix.

3107

Let z(s) be a curve in R3 parameterized by arc-length. Assume that ~ ( s ) # 0 and k’(s) # 0 for all s. Show that a necessary and sufficient condition for z(s) to lie on a sphere is that

= constant. -+-.- 1 1 P ( S )

P ( S ) + ( s ) k4(s)

(Indiana) S o h t ion.

Suppose that z(s) lies on a sphere centered at the origin. Then we may assume that Z(S) = a(s )T ( s ) + b(s )N(s ) + c ( s )B( s ) , where { T ( s ) , N ( s ) , B ( s ) } is the Fkenet frame field along z(s), and a(s ) , b ( s ) , c (s ) are suitable functions to be ascertained later. Differentiating (z(s), z(s)) = R2 with respect to s, we have ( z ( s ) , T ( s ) ) = 0, from which it follows that u(s) = 0, Vs. Differentiating ( z ( s ) , T ( s ) ) = 0 with respect to s again, we obtain

1 + k ( s ) ( 4 s ) , N ( s ) ) = 0,

which means that b(s ) = -2 t ( 8 ) . -& with respect to s, we obtain

At last, still differentiating (z(s), N ( s ) ) =

namely, k’(s) c(s ) = - k 2 ( S ) T ( S) *

Conversely, differentiating

160

with respect to s, we have

that vanishes identically because k’(s) # 0 and

Then

is a constant vector, denoted by m. Hence (z(s) - m,z ( s ) - m) = constant, namely, z(s) lies on a sphere centered at m.

3108

Let M c R3 be the torus obtained by rotating the circle ((0, y, z ) : (y - 2)’ + z’ = 1) around the z-axis, and let c ( t ) = (2cost, 2 sint, 1) (“top circle”). Is this curve a geodesic on M ? Explain without long computations.

Solution.

follows

(Indiana)

Observe that the geodesic curvature of a curve on M can be computed as

k , = f k ( t ) sin€J(t),

where k ( t ) is the curvature of the curve, and @(t) is the angle between the normal of M and the principal normal of the curve at the point corresponding to the parameter t .

Then, the curve c ( t ) is not a geodesic on M , because neither the top circle is a straight line, nor its principal normal, which is orthogonal to the z-axis, is parallel to the normal of M along c ( t ) , which is parallel to the z-axis.

161

3109

Let y : [0,1] -+ lR3 be a C" curve with 1i.l = 1 and nonvanishing curvature.

(a) Show that y lies in a plane. (b) What happens if the curvature is allowed to vanish at a point?

Assume the torsion 7 = 0.

(Indiana) Solution.

(a) 1i.l = 1 implies that the curve is parameterized by arclength s. From the Frenet formula we know that the binormal vector field of y, denoted by B , is constant. Thus, -$(y(s)B) = 0, which means that y ( s ) B = constant, that is, y lies in a plane.

(b) The vanishing curvature at point SO means ? ( S O ) = 0, i.e., SO is a stationary point of y's tangent vector field. If the point is the strict extreme value point of the tangent vector field, then it is an inflection point of the curve 7.

3110

Let f : lR -+ lR be positive and smooth. Let M be the surface in lR3 obtained by rotating the graph {(z, f(z)) : z E B} of f in the zz plane about the z axis. Characterize in t e r m of f the set of z such that && is a principal curvature of M at (z, 0, f(z)) .

Hint. Local coordinate computations are not necessary. (Indiana)

Solution.

ponding normal curvature At P = ( z p , 0, f(zp)), in the direction of the circle of latitude, the corres-

where O(zp) is the angle between the normal of M and the principal normal of the circle of latitude at P. Since every circle of latitude on M is a line of curvature, then the corresponding normal curvature Ic, is a principal curvature. Thus, Ic, = *$-J implies that cos8(zP) = f l , that is, the curve {(z,O, f(z)) : z E IR} has a tangent parallel to the z axis at P. Namely, f'(zp) = 0.

162

Besides, the meridian passing P is also a line of curvature. Its curvature Ic at P is just the other principal curvature of M . Thus, the second possible case is

Therefore, we conclude that the set of L such that f & is a principal curvature of M at (z,O, f(z)) is

3111

Let a(.) c R3 be a smooth curve parameterized by arclength. Assume that the position vector a(.) is always a linear combination of the binormal and normal vector B ( s ) , N ( s ) of a ( s ) . Show that a ( s ) does not pass through 0 E lR3.

(Indiana) Solution.

If the position vector a(.) is always a linear combination of the binormal and normal vectors, then ( T ( s ) , a(.)) 0. Integrating the obtained equality, we have ( a ( s ) , a ( s ) ) = const. Therefore, if a ( s ) passes through the origin, we will get a ( s ) 5 0, the trivial case.

3112

Let M 2 c IR3 be the cylinder z2 + y2 = 1. Suppose the curve a(.) E M 2 is parameterized by arclength.

i) If k ( s ) > 0 and T ( S ) 0, show that a ( s ) is a closed curve. (Here k , T

are curvature and torsion of a in E3.) ii) If Icg(s) 1, show that a ( s ) is a closed curve (kg is the geodesic curvature

in M).

Solution. 0 implies that the curve a is a plane

curve, whereas the hypothesis of curvature k ( s ) > 0 implies that the plane R

(Indiana)

i) The hypothesis of torsion ~ ( s )

163

where the curve a lies does not parallel the generating line of the cylinder M . Therefore, the curve a c M n x must be closed.

ii) If one develops M into a plane x, then the corresponding plane curve of a, denoted by &, has the same geodesic curvature kg(s) 3 1. Thus the

curvature of & is = dm 1 which means that it is a circle with radius 1 in x. So. a must be closed.

-

3113

Let T be a two dimensional distribution in B3 defined by

i) Show that T is not involutive. ii) Given a C” curve a ( s ) E B2 = ((z,y,O) : z,y E IR} C IR3 and

a(so) , show that there exists a unique C” curve p(s) E B3 such that ,f3’(s) E

iii) Show that if a ( s ) is a simple closed curve of length L bounding the q 3 ( $ ) , P ( S O ) = a(s0) and x(P(.)) = a(s> , where x(2, Y,.) = (z, Y, 0).

region R in B2 then @(so + L ) - p ( s 0 ) = ( O , O , &A) where A = area of R.

Solution. (Indiana)

i) That [& & + z-& = & and { &, & + z&, & are linearly independent shows that T is not involutive.

ii) Let a ( s ) = (z(s),y(s),O). By x ( p ( s ) ) = a ( s ) , we may assume that p(s) = (z(s), y(s), z ( s ) ) where z ( s ) is unknown. Then ,B’(s) E Tp(,) implies

I }

that

(:9 d2 ”> d d X

p ’ ( s ) = u(s)- + b(s) - + z(s)-

for suitable functions a(s) and b(s), i.e.,

ds ’ ds ’ d s

Hence

Thus the problem of finding p(s) reduces to solving the following initial value problem

164

When the given plane curve Q(S) is C”, the unknown function z ( s ) can be uniquely determined by

and so is the space curve P ( s ) , i.e.,

iii) If Q(S) is simple and closed, then we easily have

P(s0 + L ) - P(s0) = (z(so + L ) - z(so),y(so f L ) - Y(SO>, f ZdY)

= (O,O,fA),

where the sign is determined by the orientation of the curve a.

3114

Call a normal vector field Y along a space curve y parallel if li is always tangent to y.

i) Show that the angle through which a parallel normal vector field Y turns relative to the principal normal N along y is given by the total torsion of y, i.e.,

r ( s) ds . I” Here s and L denote arclength and the length of y, respectively.

in lR3 must vanish.

Solution. i) The hypothesis that li is always tangent to y means that (v, li) = 0, and

hence, the normal vector field v has constant norm. Therefore, without loss of generality, we may assume that (vI G 1, and v(s) = cos O(s) . N ( s ) + sinB(s) . B ( s ) , where e(s) is a smooth function globally defined on y, which measures the angle between ~ ( s ) and N ( s ) . Thus we have

ii) Show that the total torsion of any closed curve y which lies on a sphere

(Indiana)

i, = ( - s ine . N + case. B ) B +case. (-ICT - TB) + s i n e . T N = - cose k~ - ( B - .)sine. N + (4 - T ) c o s ~ . B.

165

Noting that i/ is parallel to T, we see that the above equality implies that e ( s ) = ~ ( s ) , and hence

L B(L) - q o ) = J

0

ii) For any closed curve y which lies on vector field Y of the sphere along y satisfies Therefore, we have the relation

r a

r( s)ds.

a sphere in El3, the unit normal the above mentioned hypothesis.

qS) = J, T(S)dS + q o ) .

On the other hand, the smooth function e(s) , s E [O,L] can be regarded as a lift of a certain differentiable map f : [O, L] ---f S1 into R1. The fact that y is closed means that

+)dS = e ( q - e(o) = 2n7F,

where the integer n is just the degree of the map f, i.e., n = deg f. Let po E y and y contract to po smoothly on S 2 . Thus we get a family of curves { y t } , t E [0,1]. Furthermore, we may assume that for every t E [0, l), yt overlaps with y = yo about po = yl. Also, for every ytr we have the corresponding map ft such that fo = f, fi = the constant map into PO. Because the degree of a map is homotopically invariant, finally we have

T ( S ) ~ S = 27r .deg f1 = 0. I” 3115

Let M be a surface in El3 and let P be a plane. Suppose M and P intersect orthogonally. Show that the intersection curve (parameterized by arclength) is a geodesic on M .

Solution. Let k ( s ) be the curvature of the intersection curve C = M n P. If k 5 0,

then C is naturally a geodesic on M . Otherwise, along the segment where k ( s ) # 0, the normal of M is parallel to the principal normal of C, hence the segment is also a geodesic one.

(Indiana)

166

3116

Let a(.) be a C2 curve in lR3 parameterized by arclength. Suppose that for some function f(s), a”(.) = f ( s ) a ( s ) . What can you deduce about f(s), ( Y ( S ) ?

(Indiana) Solution.

From 1 2

f ( s )a ( s )a ’ ( s ) = -d(a’(s))2 = 0

f(s)d(a(s))2 = 0. we have

If f ( s ) 0, then a”(s) If f (s) # 0, then ( ( ~ ( s ) ) ~ = const, i.e., a(s) is a spherical curve. Further-

0, i.e., a ( s ) is a straight line.

more, differentiating

with respect to s, we have

which implies that

f (s) = - k 2 ( s ) , T ( S ) = 0, k’(s) = 0.

Hence k ( s ) = const, and a ( s ) is a circle or part of it with radius 1. 0

3117

Suppose y(t) parameterizes a space curve with curvature function K ( t ) .

Define a new curve 7 by setting, for each t E lR, r(t) := cy(t/c), where c E lR is an arbitrary fixed constant. Derive the curvature function for this new curve.

Solution. (Indiana)

1 67

Therefore, from y( t ) = y ’ ( t / c ) and y”(t) = y ” ( t / c ) / c we have

3118

Let a : (a , b) + lR3 be a smooth curve with nonvanishing curvature. i) Show that if the torsion of a vanishes identically then there is a plane

?r c lR3 containing a, that is, a(t) E ?r for all t E (a ,b) . Is ?r unique? ii) Is the conclusion of i) still true if the curvature is allowed to vanish at

a single point c E (a,b)? [Of course, the torsion is not defined at c, but is assumed to be zero a t all other points.]

Solution. i) Let a be reparameterized by arclength s; namely, suppose a = a( t ( s ) ) ,

s E ( ~ 1 , s ~ ) with t ( s 1 ) = a, t ( s g ) = b. The condition k # 0 means that we can define the F’renet frame field {T, N , B } along a. Then the hypothesis T 0 implies that B is a constant vector field. Hence, $(al l?) = 0, from which follows

(a( t ( s ) ) , B ) = const = (a(t(so)), B ) .

Therefore, the plane ?r : ( p - a ( t ( s g ) ) , B ) = 0 contains the curve a. Obviously, the connectedness of a and the smoothness of the Frenet frame field imply that T is unique.

ii) If the curvature k is allowed to vanish at a single point c E (a, b) , then, according to the above discussion, a(t) , t E ( a , c ) must be on a plane T I ; and a(t) , t E (c ,b) must be on another plane a2. Of course, maybe, ?rl # x2. A counterexample is as follows. Let

(Indiana)

( t , f ( t ) , 0) if t < 0, a(t) = ( O , O , 0) if t = 0, { ( t , 0, f ( t ) ) if t > 0,

where the function f is defined by

168

3119

Let a and p be two regular curves in lR3. The curve ,f3 is called an involute of a if for all t , P( t ) lies on the line tangent to a a t a( t ) and (a’(t),,f3’(t)) = 0. Show that every involute of a generalized helix a is a plane curve. (Recall that a is a generalized helix if for some constant vector u # 0, (u,a’(s)) E const, where s is arclength for a.)

Solution. For convenience, let a and ,B be parameterized by their arclength s and

s1 respectively, and let s , s1 represent their corresponding points. Then we may assume ,B(sl) = a ( s ) + X(s)T(s ) . Differentiate the equation with respect t o s. Using the Frenet formulas and the hypothesis (T(s),Tl(sI)) = 0, we can ascertain that P(s1) = a ( s ) + (so - s)T(s). Differentiate the obtained expression for ,B successively. Noting that a being a generalized hrelix implies

(Indiana)

we obtain by straightforward calculation

i.e., the torsion of p vanishes everywhere. Therefore, p is a plane curve.

3120

Suppose that the unit normal vector to a surface M c lR3 is constant along a regular curve a c M . Deduce that in this case, a is an asymptotic (Note: A curve in a surface is called asymptotic if its acceleration is everywhere tangent to that surface.) curve, that a plane contains it, and that the Gauss curvature of M vanishes a t each point of a.

Solution.

X(ul, u2). Firstly, by the Weingarten formula, we have, along a,

(Indiana)

Let a be parameterized by arclength s , and M be locally parameterized by

(Z) = -W(a’). . duj dX o=-=-- -c?- -=-W d n ds aut

2

i J = l ds

1 69

Further, by the Gauss formula, we immediately deduce that, along a

duj duk 2 2

= kgn x a' + (W(a'),a')n = kgn x a',

where k, is geodesic curvature. Hence a" is everywhere tangent to the surface. Secondly, along a , noting that n is constant, we have d ( a ( s ) , n ) = 0. There-

fore, ( ( ~ ( s ) , n) = const = (a(so), n),

namely, ( a ( s ) - a(so) ,n) = 0, which means that the curve a is contained by a plane ( p - a ( s g ) , n) = 0.

Thirdly, from I11 - 2HII + K I = 0 it follows that along Q

K = - - + 2H(W(a'),a') = 0. ( :)2

3121

Sketch the closed regular plane curve p : [-T, T] -+ R2 having p(0) = (0,O) andP'(0) = (1,0), ifp's curvature function k ( s ) ( s = arclength) is odd, satisfies

r" Jo L(s)ds m 3 ~ 1 2 ,

and has the following graph. (Include an explanation with your sketch.) (Indiana)

Fig.3.4

170

Solution. The image of the plane curve p is a “figure eight”, as shown in Fig.3.5, which

has z-axis as its tangent line at (0,O). And y-axis is almost its “tangent” line at ( O , O ) , too. This assertion follows from

3a k ( s ) d s = J,” g d s = 6 ( ~ ) - 6(0) x T , I” where 6 ( s ) is the oriented angle formed by ,B’(s) and p’(0).

Fig.3.5

171

SECTION 2 DIFFERENTIAL GEOMETRY OF SURFACES

3201

i) Show that there exists a metric on the plane so that some geodesics are

ii) Let a(.) be a simple closed geodesic as described above and K ( z ) be simple closed curves.

the Gaussian curvature of the above metric at 2 E B2. Compute

J k. interior of Q

Here the integral is with respect to the Riemannian volume induced by the metric.

Solution. i) Let S2 = { ( 2 1 , 2 2 , 2 3 ) E B3 : 2: + xi + 22 = l}, and B2 be the

21 o 2 2 plane. For every p E R2, its coordinates with respect to the z1 and 2 2 axes are denoted by (u,v). Suppose that x : S2\{N} -+ B2 is the stereo- graphic projection from the north pole of S 2 into the plane lR2, which maps (21, ~ 2 ~ 2 3 ) E S2\{N} to (u, v) E B2. By direct calculation, we obtain

(Indiana)

2u 2v 212 + v2 - 1 u 2 + v 2 + 1 '

2 1 = 22 = 23 = u2 + v2 + 1' u2 + v2 + 1'

Then the metric of S2\{N} can be expressed by

4(du2 + dv2) ds2 = (22 + 212 + 1 ) 2 '

Now, using the pull back of ds2 by (x-')*, we can obtain a 2-dimensional Riemannian manifold (B2, (x-')*ds2). Thus, the map x : (S2 \ {N} , ds2) -+

(B2,(x-')*ds2) is an isometry. Since all great circles which do not pass through the north pole are closed geodesics on S2\{N}, then their images are simple closed geodesics on ( E 2 , (x-l)*ds2).

1 72

ii) Denote by D the simply connected domain encircled by the simple closed geodesic a. Then, using the famous Gauss-Bonnet formula, we immediately have

K ( x ) = 2TX(D) = 2T. J interior of (I

3202

Let M 2 c lR3 be a surface containing x = 0. Assume that and are ax 1 a x 2 tangent to M at z = 0, and that in that basis the Weingarten map

L = ( 4 3 -4 ” ) *

Let Q be the curve (near z = 0) obtained by intersecting M with the z1z3 plane.

i) What is the normal curvature of a at z = O? ii) What is the geodesic curvature of a at x = O? iii) What is the Gaussian curvature of M 2 at x = O? iv) Sketch the surface near x = 0. You may assume that the normal to the

surface at x = 0 is ( O , O , 1).

Solution. (Indiana)

i) By the Meusnier theorem, the normal curvature of Q at x = 0 is

ii) Since a is obtained by intersecting M with 21x3 plane, i.e., a is a normal section, then its curvature at z = 0 is L ( 0 ) = lk,(O, &)I = 4. By the relation among k ( O ) , kn(O, &) and the geodesic curvature k,(O, &)

we immediately obtain Ic , (O, &) = 0.

the Gaussian curvature of M 2 at x = 0 is K ( 0 ) = -25; or more directly, iii) Because the eigenvalues of the Weingarten map L are f 5 , we know that

K ( 0 ) = det ( y4 ) = -25.

1 73

iv) From iii), we know that x = 0 is a hyperbolic point of M 2 . Noting that the normal to the surface at z = 0 is ( O , O , 1) and k,(O, &) = 4 > 0, we sketch the surface near x = 0 as follows.

/ s

Fig.3.6

3203

Let M = {(x, y, z ) E E3 : z = 6 - (zz + y2)} (a paraboloid of revolution). D = {(x, y, z ) E M : z > 21, and w = yz2dx + zzdy + x2y2dz. Orient M and evaluate

L z d x A d y + d w .

(Indiana) Solution.

Choose the unit outward pointing normal as the orientation of M . Let P = {(z, y, z ) E E3 : x2 + y2 _< 4 ,z = 2) and take (0,0, -1) as its normal vector. Then, by the Stokes’ formula, we have that

l U p d w = 0.

Hence,

Therefore,

Firstly, we have

k [ 6 - (z2 + y2)]dx A dy = 12r dB 1 2 ( 6 - r2)rdr = 16n.

174

Secondly, noticing that

dw = z 2 d y A d x + 2yzd.z A d x + z d x A d y +zdz A d y + 2xy2dx A d z + 2 x 2 y d y A d z ,

we have

- ~ d w = - ~ 4 d y A d x + 2 d x A d y = - 2 J, d y A d x = - & .

Hence, r

J, z d z A d y + dw = 8T.

3204

Let S be a surface diffeomorphic to the ordinary 2-sphere. Suppose there is a C" metric of positive curvature on S for which there exist two simple closed geodesics y1 and 7 2 . Show that y1 and yz must intersect.

Solution. (Indiana)

Here we assume the Jordan curve theorem. If y1 and yz do not intersect, then y1 and 72 encircle a domain D such

that the Euler characteristic x ( D ) = 0 and d D = 71 uyz. Applying the global Gauss-Bonnet formula to the domain D c S and noting that y1 and y2 are geodesics of S, we have

K d c = 27rx(D) = 0,

which contradicts the assumption of positive curvature.

3205

Let D be the disk { ( x , y ) E lR2 : x2 + y2 < i} and let 0 = (0,O) E D. Let ( z 1 , x 2 ) = ( r , 0 ) be the usual polar coordinates and define a metric on D\{O} by

where h(r, 0) = r2[1 - 2r2 + r4 sin2 01'.

175

(a) Prove that this metric extends to a smooth metric on D. Hint. Use a smooth coordinate system. (b) Show that line segments in D which pass through the origin (suitably

parameterized) are geodesics.

Solution. (a) Set

(Indiana)

x = rCOS8, { y = r sin 8.

Then = r. So, on D\{O}, we can choose (x, y) as the coordinates. From

dx = cos Bdr - r sin 8d8 { dy = sin 8dr + T cos 8d8

dr = cos dx + sin 8dy { d8 = (cos 8dy - T sin adz).

follows

Hence the metric of D\{O} may be rewritten as

ds2 = dr2 + h(r,8)de2 = [cos2 8 + (1 - 2r2 + 4r4 sin2 19)~ sin2 8)dz2

+2 cos 8 sin 8[l - (1 - 2r2 + 4r4 sin2 8)2]dxdy

+[sin2 8 + cos2 8( 1 - 2r2 + 4r4 sin2 8)2]dy2

giidx2 + 2lii2dxdy + li22.d~~. - =

- When r + 0, we see that &I = 1 + o(r ) , g12 = o(T), Z22 = 1 + o(r) , and

%%i az = o( l), $& = o( 1). Therefore, this metric can extend to a smooth metric on D.

(b) For any line segment I in D, we can express 2\{O} as the union of Il = { ( r , e ) : 0 < r < :,8 = O 0 } and l 2 = {(r ,8) : 0 < T < :,8 = B0 + T } .

Using ds2 = dr2 + h(r,8)d82, we know that both l1 and 12 are geodesics in D\{O}. If we use the extended metric, the geodesic curvature k , of I must be a continuous function of the coordinates. Because 11 and 12 are geodesics by either metric, we know that kgIo = 0. Hence, as a whole line segment, 1 is a geodesic, t 00.

-

176

3206

i) Let M be a surface (without boundary) in lR3. Suppose M is inside a ball of radius R and suppose a point p E M is on the boundary sphere. Show that the Gauss curvature of M at p is 2 &.

ii) Show that there is no closed minimal surface in lR3. (Indiana)

Solution. i) The surface M and the sphere S 2 , the boundary of the ball, have the

same tangent plane at p . Let a be a normal plane of M and S 2 at p . Then, using the local canonical forms of the normal section curves M n 7~ and S2 n 7~ at p , one can easily show that, a t p , the curvature of M n T is not less than that of S2 na = S1, and p is an elliptic point of M . Hence the Gauss curvature of M at p is greater than or equal to 6.

ii) Suppose that M is a closed minimal surface in lR3. Then there is a family of spheres that contain the surface M inside and have a fixed center. Ler R be the infimum of their radii. Then, there must be at least one common point of M and the sphere with radius R and centered at the fixed point. Using the result of i), one concludes that the Gauss curvature of M at p , K ( p ) 2 &-. But it contradicts the fact that , for minimal surfaces,

3207

Let H 2 = {(z, y) E lR2, y > 0) be the upper half-plane in lR2 and let H 2 have the Riemannian metric g such that

where (2, y) is the usual inner product on B2. (a) Compute the components of the Levi-Civita connection (i.e., the Chris-

toffel symbols). (b) Let V(0) = ( 0 , l ) be a tangent vector a t the point ( 0 , l ) E H 2 . Let

V(t) = (a( t ) ,b( t ) ) be the parallel transport of V(0) along the curve z = t ,

177

y = 1. Show that V ( t ) makes an angle t with the direction of the y-axis in the clockwise direction.

Hint. Write a( t ) = cosO(t), b ( t ) = sinO(t) where O ( t ) is the angle V ( t ) makes with the z-axis.


(a) From the hypothesis we have E = G = I F = 0 and hence ?I2 '

(b) For convenience, denote the curve (2, y) = ( t , 1) = (d( t ) , u2(t)) and the parallel transport vector field V ( t ) = (a ( t ) , b ( t ) ) = ( w l ( t ) , v 2 ( t ) ) . Then V ( t ) must be the solution to the following initial value problem

(vl (O) ,v-yO)) = (0 , l ) .

Noticing the above expression of rik, along the curve, we see that the problem is equivalent to

9 = b ( t ) , = -a(t)

Therefore, writing a( t ) = cos@(t), b ( t ) = sin@(t), we have immediately 0 = -t.

3208

Consider the hyperboloid S of one sheet x 2 + y2 - z2 = 4. (a) Define the Gauss curvature K of S. (b) Use the definition from part (a) to compute K at (0 ,2 ,0 ) .


(a) The Gauss curvature K of S at P is defined by K = k lk2 , where k l , k2

are the two principal curvatures of S a t P. (b) At P = (0,2,0), in the direction of the circle of latitude, the normal

section is a circle x 2 + y2 = 4. Thus, by taking the outer unit normal vector field as the orientation of S, the corresponding principal curvature Icl = -f;

178

whereas in the direction of the meridian, the normal section is a component of the hyperbola y2 - z2 = 4, thus the corresponding principal curvature k2 = i. Therefore, the Gauss curvature K = k l k 2 = -a . 1

3209

Calculate the total geodesic curvature of a circle of radius r on a sphere of radius R > r.

Solution. (Indiana)

By the Gauss-Bonnet formula, the total geodesic curvature is given by

where C is the given circle, and s1 is the spherical cap encircled by C.

3210

Let M be a compact surface embedded in R3, with smooth unit normal vector field v.

Show that the mapping P, : M ---f R3 defined by P,(z) = x + E Y ( X ) will immerse M provided I E ~ is the reciprocal of neither principal curvature k l or k 2 at x E M . In this case, express the principal curvatures of M , := P, (M) at P,(x) in terms of k l and k 2 .

Solution.

by the Rodriques equation we have on P,(M)

(Indiana)

If, locally, we take the lines of curvature as the parametric lines of M , then

a ax -Pc(z) = (1 - & k i ) - , i = 1,2 . dU' aui

Thus,

a a ax ax ---Pc(z) x -P,(z) = (1 - & k 1 ) ( 1 - E k 2 ) - x - # 0 d U 1 au2 d u l au2

179

provided that I E ~ is the reciprocal of neither k1 or k2 at x E M , which shows that the mapping P, immerses M into R3. Furthermore, again by the Rodriques formula, we see that the u1 and u2 parametric lines are the lines of curvature in P,(M), and the corresponding principal curvatures are ~ k l - 1 and - 1, respectively.

3211

Let X : lR -+ lR be a smooth function with compact support, and consider the Riemannian surface obtained by equipping lR2 with a metric g of the form

g(v, w) := ex(Y)(v, w)

for v , w E T(,,,)lR2. Assuming the Gauss curvature K of this metric is everywhere non-negative, use the Gauss-Bonnet Theorem to deduce that in fact, the surface is flat.

Solution. Using the Descartes coordinates (x,y) in lR2, we can express the metric

of the Riemannian surface k2 by ds2 = ex(y)(dx2 + dy2), which is obviously conformal to that of the Euclidean plane B2. Take a positive number a and consider a square domain D in I R 2 , whose vertices are A(-a, -a) , B ( a , -a) , C(a, -a), D(-a , a ) respectively. Suppose that, in lR2, the corresponding - - - - part of 2) is 5, and the corresponding points of A, B , C , D are A, B , C , D , respectively.

Then, by the Liouville formula, we can see that the geodesic curvatures of the segments of coordinate curves i g and EE are

(Indiana)

-

O=O,y= -a sin 6'

1 d l n G coso+ -- do 1 a l n E 2 d E a Y 2 d z ax

O=n,y=a sin 0

1 d l n G case + -- do 1 d l n E 2 a ax

k,lco = (-- -- ds 2~ a Y

whereas both and 52 are geodesics of k2. Applying the Gauss-Bonnet

180

formula to 5 c IR2, we have

where the Gauss curvature

Noting the area element along i g and E E , from

d a = e’dxdy and the line element ds2 = ex(y)dz2 the above integral equality we can obtain

Because the number a is arbitrarily chosen, letting a -+ 0, it leads to

Besides, by the hypothesis K 2 0, we have

d2X - < 0 ; dy2 -

namely is a decreasing function. Thus, ’ dY

y=-a y = a

which combining the above equality shows that % = 0 in B1. Therefore,

K 0, i.e., R is flat. - 2

3212

Let M 2 c R3 be a smooth compact surface such that M 2 c ((2, y , z ) : z 2 0). Assume that M 2 n ((2, y, z ) : z = 0) is a smooth curve a(s ) , parameterized by arclength.

i) Show that a(.) is an asymptotic curve on M 2 . ii) Show that a’(s) is always a principal direction.

181

iii) Let's now drop the assumption that M 2 n ((2, y, z ) : t = 0) is a curve. What kind of a set could M 2 n {(z, y, t) : z = 0) be?

Solution. i) The hypotheses imply that along the curve a ( s ) , the plane {(z,y,z) :

z = 0) is a tangent plane of the surface M 2 . Thus, the unit normal vector field n ( s ) of M 2 is constant along a(s) . Therefore, along a ( s ) , = 0, which means that the tangent vector of a ( s ) for every s is an asymptotic direction. Hence, a ( s ) is an asymptotic curve on M 2 .

= -O-a ' ( s ) , from which the Rodriques equation says that a'(s) is always a principal direction with the principal curvature 0.

iii) If we drop the assumption that M 2 n {(zl yl z ) : t = 0) is a curve, then the set M 2 n {(z,y,z) : z = 0) consists of elliptic or parabolic points of M 2 , a t which the plane {(z, y, z ) : z = 0) is tangent to M 2 .

(Indiana)

ii) Noticing the above fact, we see that along a ( s ) ,

3213

(a) Construct an example of a non-compact C" surface in R3 with a

(b) Show that this is not possible if the surface is compact. sequence of closed geodesics {c.,} such that length (ri) -+ 0.

(Indiana) Solution.

(a) Consider the following surface S of revolution generated by a C" vi- brating curve C, illustrated in Fig.3.7, rotating around the z-axis which is the asymptote of the curve. Let Pi denote the points where C has horizontal tangents. Then, on S, Pi draw closed geodesics c.i, and length (ci) + 0.

X

Fig. 3.7

(b) Suppose S is a compact surface. If there exist closed geodesics c.i such

182

that length (ui) + 0, then by the Gauss-Bonnet formula we have

where a, denotes the domain encircled by u,. Because S is compact, when i is sufficiently large, we may suppose R; is simply connected. Thus, setting i --+ 00 in the above equality, we come t o a contradictory result 0 = 2 ~ .

3214

Let ( r ( s ) ,O , z ( s ) ) be a unit speed curve in B3 with T ( S ) > 0. Consider the surface of revolution (s, 0 ) + ( ~ ( s ) COSO, T ( S ) sin 8, z ( s ) ) . On this surface compute the covariant derivatives V+& and V A & in terms of & and 6.

(Indiana) Solution.

86

The direct computation gives

I = ds2 + r2(s)d02 = Eds2 + Gd02.

Denoting ( s ,e ) = (ul , u2) , we have

a = - T ( S ) T ’ ( S ) - ,

a - l a a GI 8 G2 vs- - r2,%+r2 -=---+-- 8 9 ae 2 2 a e 2 ~ a ~ 2 ~ a e a s

a a a E~ a G~ a - 8 as 8 s 21ae 2~ as 2 ~ a e - r ( s ) ae’

v8- = rg l -+y2 - +-- ___

where r;, denotes the Christoffel symbol ( i , j , Ic = 1,2).

3215

Let M 2 c R3 be an embedded compact surface of genus 2 1. Show that

(Indiana) the Gauss curvature of M must vanish somewhere on M .

Solution. By the Gauss-Bonnet formula, we have

Kda = 2 r x ( M Z ) = 4 ~ ( 1 - 9 ) _< 0. J I,.

183

Firstly, we claim that because of the compactness of M 2 , there must be a point P with K(P) > 0; hence, by the continuity, there exists a domain U of P such that KIu > 0. Secondly, we show that there exists another point Q with K(Q) < 0. Otherwise, we would have

a contradiction. Finally, by the connectedness, the continuous function K must vanish somewhere on M 2 .

3210

Consider the torus-of-revolution T obtained by rotating the circle ( z - u ) ~ + z2 = r2 around the z-axis:

T = {(z, y, Z) : (z2 + y2 + z2 + u2 - T ~ ) ~ - 4u2(z2 + y2) = 0).

Parameterize this torus, compute its Gauss curvature function K, and verify that sT KdA = 0 by explicit calculation.

Solution. (Indiana)

Thus obtained torus-of-revolution T can be parameterized by

X(u,w) = ( (u+rcosu)cosw,(a+rcosu)sinw,rs inu) , 0 < u,w < 27r.

A straightforward computation gives the coefficients of its first and second fundamental forms

E = r 2 , F = 0, G = (u + T C O S U ) ~ ;

L = T , M = 0, N = C O S U . (u+ TCOSU).

LN - M~ Therefore, its Gauss curvature function

cos u - - K = EG - F2 T ( U + TCOSU) '

Noting that the area element on T is

dA = JEG-F2dudv = T(U + TCOS u)dudv,

we have immediately

KdA = 12T dv 12' cos udu = 0.

184

3217

Let “1 2 (; 2 1 z ( t ) = -cos(t), -sin(t), - , 0 5 t < 27r

be a curve on S2 c Ill3. Let X o = & E T(L fils2. Compute the parallel translation of Xo along z( t ) .

Solution. Consider the cone that is tangent to the unit sphere S2 along the curve

z. This cone minus one generator is isometric to an open set D c lR2 given in polar coordinates by 0 < p < +m, 0 < 8 < AT. Because the parallel translation in the plane coincides with the normal Euclidean one, we obtain the result that, for a displacement t of a moving point p along z starting from (i, 0, q) (corresponding to the central angle 8 = $t in the domain D), the

2 1 7 a

(Indiana)

oriented angle formed by the tangent vector vector X ( t ) of XO is given by 27r - 8 = 27r -

with the parallel translation

3218

Show that for any Riemannian metric on S2 with IKI 5 1, where K is the Gauss curvature, the area of S2 is not less than 47r.

Solution. (Indiana)

By the Gauss-Bonnet formula, for any Riemannian metric on S 2 , we have

J,, Kdv = 2 x x ( S 2 ) = 47r.

Thus, if IKI 5 1, then

47r 5 Is IKldv 5 Area ( S 2 ) . 2

3219

Define “geodesics”, and characterize (with proof): i) All geodesics on the unit sphere S2 c IR3.

185

ii) All geodesics on the surface ((2, y, z ) E R3 : z2 + y2 E 1). ( Indiana)

Solution.

we have Let a ( s ) be a curve on a surface M parameterized by arclength. Along a ,

d ( S ) = Ic(s)N(s)

where u1,u2 are local coordinates on M , n is the unit normal vector field of M along a , and kg is the geodesic curvature of a. Then, a is a geodesic of M if and only if along a, k, E 0 or

d2ui . d u j d u k ds2 ds ds - +Cr*. -- = o i = i , 2 .

j,k

i) On the unit sphere, every great circle is a geodesic, because along a, the principal normal vector N is parallel to n. On the contrary, owing t o the uniqueness of the initial value problem

every geodesic on M , that starts from a given point and is tangent t o a given direction, must be a great circle.

ii) Since geodesics are intrinsic objects, then if we develop the cylinder to a plane, every geodesic must become a straight line. Therefore, every geodesic on the cylinder must be a helix, or a circle of latitude, or a straight generating line.

3220

Give an example (e.g., draw a picture) to show that a connected surface can have two points which are not jointed by any geodesic. What is the usual topological hypothesis that prevents this problem?

( Indiana)

186

Solution. Let K be a plane and p , y two points in K. Let T be an interior point of the

line segment p y . Then the surface K\{T} is a case in point. “Completeness” is the usual topological hypothesis that prevents this problem.

3221

Define “minimal surface in R3”, and prove that the catenoid, obtained by rotating the graph of y = cosh(z) around the z-axis in R3, is minimal.

Solution. (Indiana)

A minimal surface is a surface with mean curvature

= 0. 1 E N - 2 F M + G L 2 E G - F 2

H = -

The straightforward calculation shows that the catenoid

X ( Z , 8) = (z, cosh(t) cos(8), cosh(t) sin(8))

is a minimal surface in R3.

3222

The “Clifford” torus in S3 can be parameterized using charts of the form

1 X ( u , w) = -(cos u, sin u, cos w, sin w), Jz

where u and v are constrained to lie within intervals length < 2 ~ .

the indicated type. i) Compute the metric [gjj] on the Clifford torus for a coordinate chart of

ii) Figure out the Clifford torus’ Gauss curvature function. Hint. Calculation is not necessary here. iii) Deduce from the result of ii) that the Euler characteristic of a torus is

zero.

Solution. (Indiana)

i) From

1 1 x - -(-sinu,cosu,O,O), X - -(0,0,-sinv,cosv) “ - A “ -a

1 87

we easily have 1 2 911 = 922 = -, g12 = 921 = 0.

ii) The Gauss equation implies that the Clifford torus’ Gauss curvature

iii) Using the Gauss-Bonnet formula, we see that the Euler characteristic function K 0.

of a torus T

3223

Let f : IR2 + lR be a smooth function with a critical point (e.g., a minimum

i) Show that the principal curvatures of the graph or maximum) at the origin 2 1 = 2 2 = 0.

z = f ( Z 1 , 2 2 )

at ( O , O , f ( O , O ) ) E R3 are the same as the eigenvalues of the Hessian matrix [a2 f /az ;a2 j ] at ( O , O ) .

ii) Show that lR3 contains no compact embedded surface with strictly negative Gauss curvature at all its points.

Hint. Look at the “lowest” point on the surface. (Indiana)

Solution.

straightforward calculation, we have i) The hypothesis means that $ and vanish at (0,O). Then, by

Therefore, the principal curvatures of the graph

z = f (Z1, .2)

at ( O , O , f ( 0 , O ) ) E B3 are the roots of the following equation

188

ii) Let p ( z 0 , yo, zo) be the “lowest” point on the surface. (The existence of such a point follows from the compactness of the surface.) Since the surface is above the plane T : z = zo and p is the common point of R and the surface, then T is the tangent plane of the surface a t p . By observing the normal section at p , it is easy to conclude that any normal curvature of the surface at p is greater than or equal to zero, if we take ( O , O , 1) as the unit normal of the surface at p. Thus the Gauss curvature of the surface a t p is not less than zero.

3224

Let M be a 2-dimensional manifold smoothly embedded in lR3 with unit normal n.

a) Prove that for each p E M there exist an open neighborhood Up of p in lR3 and a smooth function F : Up -+ B2 such that F-l(O) = Up n M .

b) Find F if M is the graph of a smooth function f : lR2 + lR. (Indiana)

Solution. Since the inclusion map M + lR3 is an embedding, for each coordinate

neighborhood V c M of p E M , there exists a neighborhood U of p in R3, such that V = U M . Using the local coordinates (u, v ) for V and (z, y, z ) for U, we can express this embedding as ( u , ~ ) -+ ( z ( u , ~ ) , y ( u , v ) , z ( ~ , v ) ) . Noticing that

we know from the implicit function theorem that there exists a neighborhood V, c V of p , such that on V,,

2 = 2 ( u , v ) { Y = Y(U,

{ 2) = v ( 2 , y ) .

has smooth inverse = 4 Z l Y )

Then, we can find a neighborhood Up c U of p in lR3, such that V, = Up n M , and in terms of the local coordinates (z,y), the embedding can be expressed by (2, Y) - (z, Y, f (2 , Y)), where f(z, Y) = z (u (2 , Y>, ~ ( 2 , Y)).

Therefore, setting F : Up ---t R3 by F (z ,y , z ) = z - f ( z , y ) , we have F - ~ ( o ) = up n M .

1 89

b) If M is the graph of a smooth function f : R2 -+ lR, then for each p E M , we can take Up = R3, and F : R3 -+ R defined by F (z ,y , z ) = z - f (z ,y) .

3225

Let M be a 2-dimensional manifold smoothly embedded in R3 with unit normal n. Assume that M is the boundary of a bounded convex open set. Assume that n is the exterior normal and that the Gauss curvature K of M is everywhere positive. [Recall that S C R3 is defined to be convex if for each two points of S the line segment joining these points lies in S. You may use without proof the fact that M lies on one side of each of its tangent planes.]

a) Define the Gauss (or sphere) map

7) : M --$ s2 = {(z, y, z ) : x2 + y2 + z2 = 1).

b) Prove that 7) is one-to-one. c) Assuming that 71 is one-to-one, prove that 7) is a diffeomorphism onto

d) Show that S2.

J, K(P)dP = 4?T.

(You may assume b) and c) if you wish.)

Solution. a) For each p E M , define q ( p ) as the end point of the unit exterior normal

n(p) after parallel translating it to the origin of E3. b) and c). We first prove that q : M .+ S2 is a local diffeomorphism.

For each p E M , there exist a coordinate neighborhood C c M of p and a coordinate map h : C -+ U c B2, which is a diffeomorphism, such that on U the Gauss map has the following expression

(Indiana)

q 0 h-'(u, .> = (4% .),P(., 211, r(., .>> = n(u, v),

where n(u, v) is the unit normal at X ( u , v) E C C M .

Jacobi matrix Since nu x n,, = K X , x Xu and K > 0, Xu x Xu # 0, the rank of the

ffu ffv

rank ( /3: ) = 2.

190

Thus, the implicit function theorem says that 7 o h-l has a smooth inverse on a neighborhood of h(p ) (which may be smaller than U ) ; hence 7oh- l is a local diffeomorphism. Therefore, in the neighborhood of h ( p ) , q = (7 o h-') o h is a diffeomorphism. Thus, on M , 7 is a local diffeomorphism.

Next, we show that q : M -+ S2 is surjective. Since 7 : M -+ S 2 is a local diffeomorphism, q is an open map. Besides, because M is compact and S2 is a Hausdorff space, 7 is also a closed map. Thus q ( M ) is an open, closed and non-empty set of S 2 . The connectedness of S2 implies that q ( M ) = S2.

Now we prove that q : M ----t S2 is globally one-to-one by leading to a contradiction. Suppose that there are two distinct points P, Q E M such that q ( P ) = q(Q). From the above argument we know that there are neighborhoods C p , CQ of P,Q respectively, such that 7Ic, : C p - 7(Cp) ,q lc , : CQ -+

~ ( C Q ) are diffeomorphisrns. Because P # Q, in M we can choose C ~ , C Q so small that C p n CQ = 0. Now take the inverse images of ~ ( C P ) nq(CQ) under q l ~ , and qlcQ, respectively. Namely, set

u = ( v I c p ) - 1 ( 7 ( C P ) n 7(CQ)),

v = (vIc*)-'(7(cP) " rl(CQ))-

Thus, u n v = 0 and q ( U ) = q(V), which implies q(M\U) = S2. On the other hand, it is easy to show that M is compact, connected and oriented. Then the Gauss-Bonnet formula gives

Kda = 4n, (by noting K > 0), 1, da = I, where da, fi have local expressions

Hence

Since 5, Kda > 0, we arrive at a contradiction. In the end, noticing that differentiability is a local property, we see that

the globally one-to-one, surjective local diffeomorphism is naturally a global diffeomorphism.

191

Remark. In fact, this problem is the famous Hadamard theorem. Using

Firstly, as the above, show that q : M --t S2 is a local diffeomorphism. Since M is compact and S2 is connected, then the local diffeomorphism

q : M -+ S2 is also a covering map. Further, because S2 is simply connected, and M c R3 is connected and

hence path connected, we know that the covering map q must be a homeomorphism and hence a global diffeomorphism.

the theory of covering map, we can simplify its proof as follows.

3220

Let M be a 2-dimensional manifold smoothly embedded in R3 with unit

a) Explain what is meant by intrinsic and extrinsic quantities on M . b) Are the principal curvatures intrinsic? c) Discuss why the covariant derivative on M , defined using the covariant

d) Assuming c), discuss why the Gauss curvature of M is intrinsic.

normal n.

derivative on R3, is intrinsic.

(Indiana) Solution.

a) The terminology “intrinsic quantities” means those geometrical quantities that are definable only by the first fundamental form of M and its derivatives. Otherwise, they are called “extrinsic quantities”. In other words, intrinsic quantities are those that are invariant under isometric correspondence, but extrinsic ones are not.

b) The principal curvatures axe not intrinsic; they are extrinsic. For example, consider a plane and a cylinder.

c) Although the covariant derivative on M is defined by using the covariant derivative on R3, the last local expression of the covariant derivative of M involves the tangent vector field and the Christoffel symbols of M . Therefore, it is intrinsic.

d) For each p E M , let C be a simple closed curve encircling a simply connected domain D where p lies. Let Aw denote the angle variance caused by a tangent vector after parallel translating it around C once. Using the Hopf’s rotation index theorem and the Gauss-Bonnet formula, we can show that the Gauss curvature of M at p can be expressed by

Aw K ( p ) = lim -.

D-+P J so do

192

Since parallel translation is intrinsic, then the Gauss curvature is intrinsic, too.

3227

By revolving the curve y sketched below around the z-axis, we get a surface of revolution M 2 c R3. Compute JM KdA, where K is the Gauss curvature and dA the area form, on M . (Make sure to justify your answer.)

(Indiana)

z I

I

Fig.3.8

Solution. The surface M 2 of revolution generated by the curve y is homeomorphic

to a section of a cylindrical surface. Thus the Euler characteristic number x ( M 2 ) = 0. Besides, the boundary of M 2 consists of two circles which are just geodesics of M 2 , because along these circles the normal vector of M 2 is parallel to the principal normal vector of the two circles respectively. Therefore, by the famous Gauss-Bonnet formula, we have immediately

KdA = 0.

3228

A surface C2 immersed in a Riemannian manifold N is said to be ruled if it can be parameterized near any point by a mapping X : (0, 1)2 -+ N such that for each fixed wo E (0, l), the u-parameter curve X ( u , w o ) is a geodesic in N.

a) When n = 3, show that the Gauss curvature of a ruled surface in Rfl is nowhere positive.

b) Show it for arbitrary n. (Indiana)

193

Solution. a) Since every geodesic in N = R3 is a straight line, then C2 can be

characterized locally by X ( u , w) = a(w)+(u- i ) l ( w ) , where Z(v) is the direction of the geodesic corresponding to w E ( 0 , l ) . Thus, through computation, we can easily deduce that the first coefficient of the second fundamental form of C2

d2 X L = (n, -) f 0 , d U 2

which shows that the Gauss curvature of the surface

- < o . - - LN - iw2 K = EG- F 2 EG- F 2 -

b) Similar to the above, suppose C2 can be characterized locally by

where, for convenience, we may assume that v E ( 0 , l ) is the arclength of the curve a ( v ) , 1Z(v)1 SE 1, and (a’(u),l(w)) F 0. Then, by a routine work, we see that the first fundamental form of C2 is

ds2 = d U 2 + (1 + 2u(a’(w), Z’(v)) + U2II’ (V)12)dV2.

Since the Gauss equation says

K = -* d z ’

it suffices to show that 2 0. However, we have

Noting that

1l’(v)12 - (a’(v),Z’(w))2 = la’(w) x I’(w)I2 2 0,

we obtain the desired result.

194

SECTION 3 DIFFERENTIAL GEOMETRY OF MANIFOLD

3301

Let M" be a Riemannian manifold and R a 2-plane in T, M. i) Let { X , Y } be an orthonormal basis of R . Use this basis to define the

sectional curvature K ( R ) and show that it depends only on R and not the particular basis chosen.

ii) Define the Riemann curvature tensor in terms of covariant differentiation. Explain why it is a tensor.

iii) Recall that if M" c En+' is a hypersurface, then

R ( X , Y ) Z = ( L Y , Z ) L X - ( L X , Z ) L Y ,

where L is the Weingarten map. Using this or any other valid method, compute all sectional curvatures of the sphere {z E B4 : 121 = 3).

Solution. i) The sectional curvature is defined by K ( R ) = - R ( X , Y , X , Y ) , where

R( ) is the Riemannian curvature tensor field of M". If ( 2 , Y ) is another orthonormal basis of R , then 2 = aX + by, ? = cX + d Y . Noticing that

( 1 ) is an orthogonal matrix, and R( ) is a 4th order covariant tensor

field, we can easily have R ( X , Y , X , Y ) = R ( X , Y , X , Y ) , which proves that K ( R ) depends only on R .

ii) For any vector fields X , Y , 2, W on M " , define the Riemannian curvature operator by R ( X , Y)Z = VxVyZ - VyVxZ - V[X,Y]Z and the Riemannian curvature tensor field by R ( X , Y, 2, W ) = ( R ( X , Y ) Z , W ) , respectively. Then for any C" function f on M " , using the properties of covariant differentiation and of the inner product, we can conclude from straighforward computation that

(Indiana)

- - - -

R(fX, y, z, W ) = R ( X , fY, z, W ) = R ( X , y, fZ, W ) R ( X , Y , Z , fW) = f R ( X , Y , Z , W ) . =

Thus, in terms of local coordinate frame field, one can show that V p E M", R ( X , Y, 2, W)I, is dependent only on X,, Y,, Z,, W, E T'(M"). Therefore,

195

R( ) I p is a well-defined 4th order multilinear function. Besides, the inner product and covariant differentiation are all C". Hence, R( ) thus defined is a C" tensor field.

iii) For the sphere Mn = {z E lR4,1zl = 3} , we can regard the position vector field 2 as the normal vector field of Mn. Let { e i } be a local orthonormal frame field about 3: (as a point) of Mn. Then by the original definition of covariant differentiation, one can easily show that V e i z = ei , where V denotes the covariant differentiation in lR4. Hence, the Weingarten map is L : X F-+

L ( X ) = -Vxg = -X/3 . Therefore, if { X , Y } is any orthonormal basis of an arbitrary 2-plane T in T p ( M n ) ,

- - -

K(7r) = - ( R ( X , Y ) X , Y ) = ( L X , X ) ( L Y , Y ) - ( L Y , X ) ( L X , Y ) = 1/9.

3302

Let C = {z E lR3 : 0 5 3ci 5 1) be the unit cube in B3. Suppose F : lR3 + lR4 is a 1-1 C" immersion in some neighborhood of C. The image F(C) is then a compact Riemannian submanifold of lR4 with boundary and therefore has a volume. Justify the following formula:

196

Let Au be the volume of the parallelopiped spanned by the vectors PzP;, P3Pl and in lR4. Then

is an infinitesimal when Ax1 -+ 0, Ax2 -+ 0, Ax3 ---f 0. We take the principal part of the infinitesimal Au as the volume element, namely, the volume element of F (C) is

Hence the desired formula follows immediately.

denoting F(x) = F(z1,22,23) = (yl, y2, y3, ~ 4 ) , we have that Furthermore, if F(x) = ((1 XI)^, (1 + ~ 2 ) ~ , (1 + ~ 3 ) ~ , (2 + ~ 3 ) ~ ) , then, by

F, (&) = 2 ( 1 + ~ 1 ) - - , a aYl a

= 2(1+x2)-, a Y 2

dY3 dY4 = 2 ( 1 + ~ 3 ) - + 3 ( 2 + ~ 3 ) a 2 8 -.

a a ayi %a

Since - - and

form an orthonormal frame, we obtain that

3303

There is no submersion from S3 into lR2. (Indiana)

1 97

Solution.

(2‘))” = 1). Observe the following three vector fields Let S3 c E4 be defined by { ( x ’ , x 2 , x 3 , x4) E E4; ( x ’ ) ~ + ( x ~ ) ~ + ( x 3 ) ) ” +

where (xl, x 2 , z3, z‘) E S 3 . It is easy to see that they are nowhere-vanishing tangent vector fields on S3. Since ( x ’ ) ~ + (z2)>” + (2’))” + ( x ~ ) ~ = 1, then, without loss of generality, we may assume x4 # 0. Hence, from the fact that

det ( x4 x3 -x2 ) = ( x ‘ ) ~ + x4(x3)’ + ~ ~ ( 2 ’ ) ~ + ~ ~ ( 2 ’ ) ~ = x4 # 0,

we know that X I , X2, X3 are three linearly independent vector fields. Further- more, by direct calculation, we obain

z2 -21 2 4

-23 2 4 2 1

Now we suppose that there is such a submersion a : S3 --+ El2. Then, a, : Tp(S3) -+ T T ~ p ~ ( E 2 ) is a surjection for every point p E S3 . Thus, we may assume that, for example, a*Xg, 7r*X3 are linearly independent, and 7rIT,X1 = aa,Xz + b7r,X3 a t p . Then, on the one hand,

on the other hand,

Therefore, a,X3 = -aba,Xz - b2a*X3

from which we get ab = 0 and -b2 = 1. Similarly, from [ X s , X 1 ] = 2x2 we can deduce that -a2 = 1 and ab = 0. They are all contradictory. So, there is no submersion from S3 into Ez.

198

3304

Let Mn and N k be Riemannian manifolds. Then M x N is naturally a Riemannian manifold with the product metric. If XI, . . . , x, are local coordinates on M and y1, * * * , yk are local coordinates on N , the product metric in local coordinates ( X I , , x, , y1, . - , y k ) looks like

i) Let X be a vector field on M x N “along” M (i.e., in local coordinates no yp and & are present) and Y be a vector field along N . Show that DxY 0.

ii) Show that a t z E M x N , some sectional curvatures always vanish. (e.g., product manifolds in the product metric never have strictly positive curvature.)

Solution. i) Using the properties of the Riemannian connection on M x N , we only

need to verify D e & = 0. In fact, observing that the inverse matrix of g is

(Indiana)

of the form

and the first class of Christoffel symbols of M x N satisfy

we easily conclude that

Obviously, we also have D s a = 0. B y p azi

ii) By the definition of the curvature operator, using the result of i), we

This means that R (&, &) = 0, Hence, for arbitrary X along M and Y along N , using the Cm(M x N)-linearity of the curvature operator, we have

R ( X , Y, X, Y ) = 0.

Therefore, the sectional curvature determined by X and Y always vanishes. Obviously, here X may involve yp, and Y may involve 2,.

3305

Let F : M --t N be smooth, X and Y be smooth vector fields on M and N , respectively, and assume F,X = Y ; that is, that

a) Let w be a smooth 1-form on N . Define the Lie derivative Lyw of w with respect to Y . (If you use local coordinates, you must verify independence of choice of coordinates.)

b) Prove that F*(Lyw) = L x ( F * w ) .

c) Let Z be a smooth vector field on M such that F,Z = W , where W is a smooth vector field on N . Show that L y W = F , ( L x Z ) .

Solution. a) Lyw is defined by Lyw = d(Y [w)+Y Ldw, where the symbol ''Y denotes

the interior product of a vector field with a form, for example, if R is a pform, then Y [R is a ( p - 1)-form defined by

(Indiana)

Thus, Lyw is a well-defined 1-form on N .

a smooth vector field on N . Then we have b) Let Z be a smooth vector field on it4 such that F,Z = W , where W is

200

c) Noticing that L y W = [Y,W] and [F,X,F,Z] = F,[X ,Z] , we immediately obtain LyW = F,(LxZ) .

3306

Let T , 8, z be the usual cylindrical coordinates in R3. Let

w = [ ~ T Z sin 8+3z2r cos 8+5r2 sin2 8]d8+[-2r cos 8+6zr sin 8]dz+[4zr2 sin 8]dr.

Let the curve y be given in rectangular coordinates by

y ( t ) = ( z ( t ) , y ( t ) , z ( t ) ) = (cost,sint,4sin5t+sin2tcosst), o 5 t 5 27F.

Evaluate s,, w . (Indiana)

Solution. Let a function f be defined by

f(r, 8, Z ) = -2rz cOs 8 + 3z2r sin 8, (T , 8, .> E B ~ .

Then f is a C” function in lR3 and we have

df = [ ~ T Z s i n 8 + 3 z 2 r ~ ~ ~ 8 ] d 8 + [ - 2 r ~ ~ ~ 8 + 6 ~ ~ s i n 8 ] d z + [ - 2 z cos8+3z2 sin81dr.

It is easy to see that f I-, is also a C” function, and by the invariance of the form of first order differentiation, the above expression of df is an exact l-form on the closed curve y. Noticing that y is a closed curve, we have, by using the Stokes’ theorem,

w = I (w - d f ) = 5r2 sin2 8d8 + (4zr2 sin 8 + 22 cos 8 - 3z2 sin 8)dr I Without loss of generality, we may assume 0 5 8 < 27r and T = 1. Then

i w = l r 5 s i n 2 8 d 8 = 5 d8 = 57~.

201

3307

Let M be a C" Riemannian manifold. Assume the theorem that there is a unique C" mapping V : X ( M ) x X ( M ) + X ( M ) denoted by V : ( X , Y ) -+ VXY which has the following linearity properties: For all f , g E Cm(M) and X , X ' , Y, Y' E X ( M ) , we have

VfX+gX, Y = f ( v x y ) + S ( V x , Y > ,

[ X , Y ] = VXY - VYX, V x ( f Y + gY') = f V x Y + gvxy ' + ( X f ) Y + (Xg)Y',

X(Y, Y ' ) = (VXY, Y ' ) + (Y, VXY').

a) Suppose a : [ q b ] --+ M is a smooth curve. Define what it means for a vector field Y on a to be parallel along a. Derive the differential equations that must be satisfied if Y is parallel along a.

b) Let X , Y E X ( M ) . Let p E M and let a : [0, b] ---t M be an integral curve of X such that a(0) = p , da/dt = X(a(t)) . Show that

where P,,o,t : T,(o)M + T,(t)M is the parallel transport along a.

Solution. a) As known, thus defined mapping V is the Riemannian connection. Using

the properties of V , we can prove that V%Y is completely determined by a( t ) and Y(a( t ) ) . So V%Y is well-defined for every vector field along a.

Now we give a definition that a vector field Y on a is parallel along a, if and only if V%Y = 0, Vt E [a, b]. In order to derive the differential equations, we choose a coordinate neighborhood with local coordinates {xa}. Then

(Indiana)

Denoting

202

we have, by the properties of V ,

Therefore, the desired differential equations are

b) Choose a basis { e l , - - - , e , } of T,(M), where n = dimM. Let

Since P,,o,t is a parallel isomorphism for every t , then { e l ( t ) , . . , e , ( t ) } is a basis of Ta( t ) (M) . Hence we can denote

Then, by the properties of the connection V and by the fact that e i ( t ) is parallel along a(t) , i = 1,. . + , n, we immediately have

On the other hand, because

is equivalent to

ei = p ~ h , t ( e i ( t ) ) ,

we can write

203

3308

Let M be a Riemannian manifold with the property that given any two points p , q E M the parallel transport of a vector from T p M to T,M is independent of the curve joining p and q. Prove that the curvature of M is identically zero, i.e., R ( X , Y ) Z = 0 for all X , Y , Z E X ( M ) .

Solution.

local coordinates zl, - . , zn . Let

(Indiana)

For an arbitrary point p E M , take a coordinate neighborhood D of p with

be n linearly independent vectors in T p M . Using the hypothesis that the parallel transport of a vector from T,M to T,M is independent of the curve joining p and q , then for q E D, we can transport every Vpi from TpM to T,M. Thus, we can define n linearly independent vector fields V1, . . . , V, in D. Obviously, all V;’s are well-defined, and if we transport along special coordinate curves, then

namely k wll,i(z) = 0 i, j , k = 1, * * * , n.

Now, by the Ricci identity, we have

k k vil,jrn(zC) - v i l , m j ( z ) = - C v11(z>Rfjrn(z) = 0.

I

Noting the linear independence of K ’ S , we immediately have, in D,

Rljm k = 0 k , l , j , m = l , - - . , n ,

i.e., the curvature of M is identically zero.

204

3309

Let M be an n-dimensional Riemannian manifold. Suppose there are n orthonormal vector fields XI,. . * , X, that commute with each other (i.e., [Xi, Xj] = 0; i, j = 1, . , n), show that the sectional curvature of M is identically zero.

Solution. Set

( I n d i a n a )

k

Then, from (Xi,Xj) = 6;j it follows that

(vxkxiIxj) + (xi> 'Xkx'j) = O!

i.e., T i i + rtj = 0; i, j, k = 1,. . , n. On the other hand, from [ X i , Xj] = 0 it follows that

vx;xj - VX,Xi = [Xa,Xj] = 0,

i.e., rk. - r!. = 0; i,j, k = 1,. . ,n. Namely, the Christoffel symbols are antisymmetric with respect t o k,j, and symmetric with respect to i , j . Therefore,

' 3 3%

rk. = -rJ = - r i i = rfj = rfk = -rk. = -r.. k '3 ak 3' a3

which means identically zero.

= 0; i, j , k = 1, . . . , n. Hence the sectional curvature of M is

3310

Let X be a smooth vector field on a Riemannian manifold M . The diver-

i) If M is closed (i.e., compact without boundary), show that gence of X, denoted by div(X), is defined by the function trace (VX) .

/ M div(X)dv = 0.

ii) If M is compact with boundary dM, show that

div(X)dw = 1, (X, N ) d s , J M

205

where N is the outer normal vector field of a M .

Stokes' Theorem.

S o h t ion. In fact, this problem is the famous Green theorem. The outline of the proof

is as follows. Firstly, one can show that, for example, by means of the normal coordinate

system about p E M , thus defined div(X) satisfies div(X)QM = d(i(X)Qm), where Q M is the volume form of M and i( ) is the interior product operator.

Next, for p E d M , choose an oriented orthonomal frame field about p {el,...,e,} such that, at p , el = N p . Let { w l , - - . , w n } be the dual frame field of {el,. . , en}. Then the volume elements of M and d M are respectively

Hint. Consider w E h l ( M ) defined by w(Y) = (X,Y) , and try to use the

( I n d i a n a )

dv = n ~ ( p ) = w ~ A * * * A w ~ , ds=RaM(p) = W 2 A * . . A w n .

Observing that, a t p ,

i(X)n, = i(X)(w1 A . . . A w n )

= =

w1(x)w2 A . . A w n + (terms involve w ' )

(x, N ) Q ~ M + (terms involve wl),

and along d M w1 s 0, one can obtain, by Stokes' theorem,

/M div(X)dv = d(i(X)QM) = J,,(X, N ) % M = J,,IX, W d s .

If M is compact without boundary, the right hand side of the above formula vanishes naturally.

3311

Let w l , .. * , wk be one-forms. Show that {wi}fZl are linearly independent if and only if w1 A w2 A

Solution.

. A wk # 0. ( I n d i a n a )

Let wl, 1 - . , wk be defined on an n-dimensional manifold with n 2 Ic. Suppose w1 A . . . A w k # 0. If wl, 9 . . , wk are linearly dependent, then

without loss of generality, we may suppose that

k - 1 W k =Ulwl$"'+Uk-lw

206

with suitable functions al , , a k - 1 . Thus we have

W1 A * * * A WWk W1 A * * * A Wk-' A ( U l W ' + . . . + ak-1Wk-l) T 0,

which contradicts the above hypothesis.

them to a basis {W~,...,W~,W~'~,...,W~}. Thus On the contrary, if wl,. . . , w k are linearly independent, then we can extend

w1 A .. . A w k A wk+l A ... A w k # o

implies that

3312

Let M be a Riemannian manifold. Let p E M (a) Show that there exists S > 0 such that

exp, : Bb(0) c TpM -+ M

is a diffeomorphism onto its image. (b) Show that there exists E > 0 such that exp,(B,(O)) is a convex set. Hint. let d(z) = distance from z to p . Show that d2 is convex in a neigh-

borhood of p .

Solution.

that d exp, is nonsingular at p, and then the implicit function theorem.

C. Whitehead. Refer to every standard textbook on differential geometry.

(Indiana)

(a) This is just the existence of the normal neighborhood of p . Use the fact

(b) The existence of convex neighborhoods is a classic result due to J. H.

3313

Show that (i) A is a differentiable manifold.

207

(ii) A is a Lie group with the standard matrix multiplication as a product. ( I n d i a n a )

Solution. Define a map F : Gl(2,LR) + Gl(1,LR) by F ( X ) = detX. Then F is a

smooth homomorphism between Lie groups, and the rank of F is constant. Therefore, the kernel of F

kerF = F-'(l) = A

is a closed regular submanifold of GZ(2, El) and thus a Lie group.

3314

(a) Let f be a smooth function on a Riemannian manifold M. Let gradf be the vector field defined by the equation

(gradf, w ) ~ = dp f v , w E TpM.

Let (z',. . . , 2") be local coordinates around p . Find the expression for gradf in terms of z', - - , 2".

(b) For a vector field X define the divergence of X, div(X) as the trace of the operator Y + DyX where D is the Levi-Civita connection. Find the expression for the divergence of X in a local coordinate system (z', . . . , P).

(c) Use (a) and (b) to find the expression for the Laplacian A in local coordinates, where A acting on a smooth function f is defined by Af = div(Vf).

Solution. For convenience, we omit the suffix p . (a) Let

( I n d i a n a )

1 8 ax1

grad f = c a -. 1

Then, from

it follows that

k

208

Thus,

(b) Denote a x = c x i - .

a x : i

Then, by the definition of divergence, we have

Af = div(gradf) = c i k a f i

3315

Let M be a compact connected Riemannian manifold without boundary.

Hint. Use the definitions in Problem to show that div( f V f) = IV f 1 2 + f A f . Let f be a smooth function satisfying Af = 0. Show that f = const.

(Indiana) Solution.

For any function f , by straight calculation, we have

div(fVf) = lVfI2 + fAf .

Now, noticing the hypothesis of this problem, by Green’s theorem we have

J, lVf l2dW = 0,

which implies df = 0 everywhere, that is, f = const.

209

3316

Suppose F : M - N is a smooth map between differentiable manifolds, and is homotopically trivial. Show that in this case, F*w will be exact whenever w is a closed 1-form on N . (Note: F is homotopically trivial if it extends to a smooth mapping : M x [0,1] -+ N such that B(z, 0) = F ( z ) , for all z E M , while F(z, 1)

Solution. Consider the map G : M + { q } and the inclusion i : { q ) ---f N . Then the

map F : M -+ N is homotopic to the composition ioG. Furthermore, we know that F* and (i o G)* = G* o i* induce the same homomorphism

q E N (q constant) for all z.) ( I n d i a n a )

F** = ( ioG)** * . H 1 ( N , d ) - H1(M, d).

Since i * ( Z 1 ( N , d)) c Z1({q), d ) = 0, the induced homomorphism F** = (i o

G)** is a zero homomorphism, i.e., F * ( H 1 ( N , d ) ) = 0. In other words, for every w E Z 1 ( N , d ) , F*w E B 1 ( M , d ) , namely, F*w is exact.

3317

Regard lR9 as the space of all 3 x 3 matrices with real entries. Does the subset

C = {A E Rg : det(A) = 0)

form a smooth submanifold of R9?

Solution. Observe that {A E lRg : rankA 5 l), the union of all axis in Rg, is

closed. Then M := R9\{A E R9 : rankA 5 1) is an open submanifold of Rg. Define a map F : M + B1 by F(A) = det(A), A E M . Noting that dF = ( A l l , A12, A13, A21,A22, A23, A31, A32, A33) where Aij is the algebraic complement of the corresponding entry aij of A, we see that rankdF = 1 on M . Therefore, F-l(O) = {A E IR9 : det(A) = 0) is a closed regular submanifold of M . Hence, it is also a submanifold of lR9.

( I n d i a n a )

210

3318

Let

Compute ss3 w , where S3 = { x E IR4 : 1x1 = l), oriented as the boundary of the unit ball (assume standard orientation on B').


Denote D4 = { x E IR4 : 1x1 5 1). Then, by the Stokes theorem we have

= J , , 4 w = J , 4 d w

(dxl A d22 A d23 A d24 + d22 A dxl A d23 A d24 = L

+dx3 A dxl A d22 A d24) x2

dxl A d22 A d23 A d24 = v01(D4) = -. = J,. 2

3319

Prove that

is a three-dimensional submanifold of IR4. ( I n d i a n a )

Solution. Define a map F from E4\{(0, O , O , 0)) into B1 by

Then F is a C" map with ( 2 4 - 23 - 2 2 x l ) as its Jacobi matrix which has constant rank 1 on IR4\{(0, 0 ,0 , 0)). Thus

( 2 1 , 2 2 , 2 3 , 2 4 ) : rank

21 1

is a regular submanifold of R4\{(0, O , O , 0)) with dimension 3. Noting that E4\{(0, O , O , 0)) is an open submanifold of B4, we see that

is also a three-dimensional submanifold of lR4.

3320

Let vector fields XI, X2 on lR4 be defined by

a a a a ax2 ax3 axl ax4 x 2 = - + 2 2 - - . x 1 = - + 21-,

i) Is there a 2-dimensional submanifold M 2 of lR4 such that for each p E

ii) Is there a nonconstant function f in the neighborhood of 0 E lR4 such M2, Xl(P), X2(P) E T,M2?

that X1 f E 0 and X2 f

Solution.

does not satisfy the Frobenius condition.

O? (Indiana)

a i) No, there is not. The reason is that the bracket [Xl,X2] = & - asJ ii) Yes, for example, we can set f = 21x2 - (23 + 2 4 ) .

3321

Let M = { ( X , Y ) : X , Y E B3, IIXII = 1, llYll = 1, ( X , Y ) = 01. i) Show that M is a smooth compact embedded submanifold of IR6 and

ii) Show that M is orientable. explain how M can be identified with the unit tangent bundle of 5''.

(Indiana) Solution.

i) Identify lR6 with {(x, y) : x, y E lR3} and define a map F : R6 -+ IR3 by

F ( x , Y) = (fl, f2 , f3) = (11x112, llY1I2, (x, Y)).

It is easy to verify that the Jacobi matrix of the C" map F

212

has constant rank three when fl = 11z1(2 = 1, f2 = ( ( ~ ( 1 ~ = 1 and f3 = (z,y) = 0. Therefore, F-’(l , 1,O) = M is a closed embedded submanifold of B6. Besides, since II(z,y)II = (llzllz + Ilyl12)i = fi which means M is bounded in IR6, M must be compact.

Naturally, for every (z,y) E M , if we regard z E S2 and y E T,(S2) with 1)yI) = 1, then we can identify M with the unit tangent bundle of S 2 .

ii) Because S2 is orientable, S2 has a covering { ( U , , # J ~ ) } of coherently oriented coordinate neighborhoods. By using identification of M with the unit tangent bundle of S 2 , for every (z, y) E M , z has the local coordinates (ui,ui) E U,, and y is uniquely determined by the oriented angle 0, at z from 4 to the unit tangent vector y E T,(S2) . Thus, (U , x I,,& x $,) is a coordinate neighborhood of (2, y) E M , where I, = (0, - E , 0, + E ) with E

being a suitable positive real number and 4, is the map from (2, y) to 0,. If (U, x I,, 4, x 4,) n (Up x I p , #Jp x $ p ) # 0, then the transition function

has the following J acobian

au,

which means that { (Va x I, ,dol x 4,)) form a covering of coherently oriented neighborhoods. Hence, M is orientable.

3322

Let F : M 4 N be a local isometry between connected Riemannian manifolds M and N . Show that if M is complete, so is N and F is a covering map.

Solution. Because F is a local isometry, F ( M ) is open in N . If y is a limit point of

F ( M ) in N , then there is a point z E M such that there exists a geodesic in N connecting F ( z ) and y. The local isometry of F and the completeness of M imply that the above geodesic can be uniquely lifted to a geodesic in M starting from z, and the image of its end point under F must be y. Therefore, F ( M ) is also closed in N . Thus, the connectedness of N implies that F ( M ) = N , i.e., F is surjective. Besides, since F maps every geodesic of M into a geodesic of N , the Hopf-finow theorem means that N is complete, too.

Next, we show that F is a covering map. For every x’ E N , take S > 0 so small that exp,, : B’(S) -+ Bi is a diffeomorphism, where B’(6) = {v E

(Indiana)

21 3

T , f ( N ) : IvI < S} and Bi = {y’ E N : d(y ’ ,z ’ ) < 5). Since F is a local isometry, F-l(z’) is discrete. Denote F-l(z’) = {z,} c M and set Ba(S) = {v E T,-((M) : lvl < 6) and Br = {y E M : d ( y , z a ) < a}. Then, we claim that Bi is an admissible neighborhood of z’ and F is a covering map.

Firstly, we claim that F- l (Bi ) = UB?. In fact, if z E F-’(B;), then

there is a unique geodesic y : [0,1] -+ B; such that y(0) = F ( z ) and y(1) = 2’. Since F is a local isometry, there exists a geodesic y : [0,1] -+ M such that F ( T ( t ) ) = y ( t ) , V t . Hence F(y(1)) = z’ and y(1) = z, for some a. Besides, L ( y ) = L(y) < S means that z = T(0) E Br, i.e., F-l(B;) c U B r . On the

other hand, u Br c F-l(B;) is obvious. Thus the claim is proved.

0

a

0

Secondly, we say that for any a , F : B,O -+ Bi is a diffeomorphism. Since M is complete, we have the following commutative diagram

and expZe is surjective. Besides, we know that d F , exp,, are diffeomorphisms. Therefore, F o expZm = expzf odF is a diffeomorphism and hence exp,, is an immersion. So, expXm is also a diffeomorphism. Hence F = exp,, odF o (exp,.)-’ is a diffeomorphism.

Thirdly, we claim that if Q # ,8, then BF n Bf = 8. Otherwise, if there is z E B,” n B f , then there exist unique geodesics y,, yp connecting z with z,, zp respectively. Let y be the unique geodesic in Bi connecting F ( z ) and 2’. Because both F : B,“ -+ B; and F : B6p -+ B; are isometric, we have F(y,) = y = F ( y p ) . In other words, ya,yp are the lifts of y through z. The uniqueness implies ya = yp. In particular, z, = ya(l) = yp(1) = zcp, which contradicts a # p.

Let F : M -+ M be an isometry of a Riemannian manifold M . i) Show that each component of X = {z E M : F ( z ) = z} is an embedded

totally geodesic submanifold of M .

214

Hint. Use exponential coordinates. ii) Give an example in which the components of X have different dimen-

sions.

Solution. i) First, we show that X has submanifold structures. For z E X c M ,

set B(6) = {v E T,(M) : IvI < 6) and Ba = {y E M : d(z,y) < 6}, where 6 is so small that exp, : B(6) -+ Bg is a diffeomorphism. Define V = {v E T,(M) : d F ( v ) = v}. Thus, V is a subspace of T,(M). Then we claim that XnBa = exp,(VnB(6)). If this is proved, because exp,(VnB(G)) is obviously a submanifold of M , we can assert that X has submanifold structures. In order to prove the claim, we first assume that y = X n & and v E B(6) such that exp,v = y. Let y : [0,1] 4 M be the unique shortest geodesic y(t) = exp,(tv) connecting z and y. Since z,y E X , and F is an isometry, F ( y ) is also a shortest geodesic connecting F ( z ) = z and F ( y ) = y. Thus the uniqueness implies F ( y ) = y. In particular, d F ( j ( 0 ) ) = j(O), namely, d F ( v ) = v. Therefore, v E V which means that y E exp,(V n B(6)), i.e., X n Bg c exp,(V n B(6)). On the other hand, suppose that v E V n B(6) and y = exp, v. Let the geodesic y : [0,1] 4 M be defined by y ( t ) = exp,(tv). From d F ( v ) = v follows d F ( T ( 0 ) ) = j ( 0 ) . Then, that F is an isometry implies F(y) = y. In particular, F ( y ) = F(y(1)) = y(1) = y, which means that y E X n Bg, i.e., exp,(V n B(6)) c X n Ba.

Next, we show that every geodesic y : ( a , b ) -+ X parameterized by arclength is also a geodesic of M . For any SO E ( a , b ) , let <(s) be a geodesic of M such that <(so) = SO), i(so) = ? ( S O ) . Since F(<(so) ) = (‘(SO),

d F ( ( ( s 0 ) ) = ((so) and F is an isometry, then F ( < ) and < are two geodesics of M which satisfy the same initial conditions. Therefore, F ( < ) = <, i.e., < lies in X . Besides, < is naturally a geodesic of X. Thus, in a neighborhood of SO, < = y. Because so is arbitrarily chosen. y is a geodesic of M . Hence, X is totally geodesic.

(Indiana)

ii) Let

M = {(qy,.) E IR3 : y > 0 , z = 0) u {(z,y,z> E IR3 : 2 = 0,y < 0},

and F be a reflection with respect to the plane z = 0, i.e., F (z ,y , z ) = (z, y, - z ) . Then M is a 2-dimensional manifold, F is an isometry, and

x = {(2,y,z) E ~3 : > o , ~ = 01 u {(z,y,z) E 1123 : = o , ~ < o , ~ = 0).

21 5

3324

Compute the de Rham cohomology groups of the circle S'. Do so directly; Le., without citing the de Rham Theorem.

Solution. (Indiana)

since Bo(S1,d) = 0 and S' is connected, then

H*(S ' ,d ) = Zo(S',d) = {f E Coo(S1,R1) I df = 0) E B'.

For H k ( S 1 , d ) , k > 1, since there are no non-vanishing k-forms on S', we

Besides, observe that have Zk(S1, d ) = 0 and hence H k ( S 1 , d ) = 0.

Z'(S',d) = Cm(S1,A'(S1)), B1(S1,d) = {d f I f E Coo(S1,R1)}.

Let 6 be the polar coordinate characterizing S1. Then & is a non-vanishing vector field on S1. Let d6 be its dual non-vanishing 1-form on S'(Caution: Here d6 is only a formal symbol, because 6 in usual sense is not a globally well-defined function on S'), and it is not exact. For every w = g(O)d6 E C"(S', A'(S')), define a function

Because R(0) = R(27r), n is globally well-defined on S1. Hence, denoting

we see that w - Cd6 = do, i.e., w - CdB is exact. Therefore,

H1(S1,d) = Z'(S',d)/B'(S',d) = (Cd6 I C E Rl} Z R1.

3325

Let X denote a submanifold of Euclidean space En, and set

u,x B ( X , E )

:= {z + I/ : 2 E x,v E N,X, IvI < E } ,

:= {y E En : Iy - z1 < E for some z E X } .

216

Show that U,X c B ( X , E ) for all E. Show that the two are not generally equal. (Consider examples of 1-dimensional submanifold in E2.) Can you give conditions which imply equality?


Setting y = z + v immediately implies that V E X c B ( X , E ) .

Let X be an open line segment in E2. Then considering the boundary of

If X has no boundary, e.g., either compact or complete, then U,X = X , i.e., the end points, can show that V E X is a proper subset of B ( X , E ) .

B ( X , E l .

3326

Consider a Riemannian manifold ( M , g ) . Call a vector field 2 on M a

i) Show.that when 2 is Killing, we have Lzg = 0 , i.e., killing vector field if 2 generates a 1-parameter group of isometries of M .

Z ( g ( X , y>> = S(JhX, Y ) + g ( X , L Z Y ) (**)

for all vector fields X and Y on M . Here Lz denotes the Lie derivative along z.

ii) Show that the expression (**) above is equivalent to

where V denotes the Levi-Civita connection for ( M , 9 ) .


In local coordinates (zl, - a , z m ) of ( M , g ) , let the 1-parameter group of

isornetries generated by a vector field 2 = C zi& be expressed by 2 =

2(z1,

m

i = l , z m ; t ) := 2 ( z , t ) such that

m m

i , j = l k , l = l

where g , j ( z ) = g (&, &), and 2 = 2(z , t ) satisfies

217

Differentiating the obtained equality with respect to t and then setting t = 0, we obtain

which can be written as k k

g k j z , i + S i k z j 0, or equivalently

g ( V a Z , - lj) $ 9 ( l i - , V e Z er3 ) = o .

From this follows what we desire in ii). Noting that the Levi-Civita connection V satisfies

Z ( g ( X , Y)) = g ( V z X , Y ) + g ( X , V Z Y )

and the Lie derivative satisfies L z X = [Z, X I , we easily obtain (w).

3327

Let M 2 be a connected Riemannian manifold and X , Y complete vector fields on M . Assume that the flows Xt and Yt are isometries of M for all t .

i) Show that the integral curves of X are curves of constant geodesic curvature.

ii) Assume that X and Y are linearly independent at all 2 E M 2 and that their flows commute Xt o Y, = Y, o X t . Conclude that ( X , X ) , ( X , Y ) and (Y ,Y ) are constant on M .

Solution. i) For every p E M , take a local coordinate neighborhood about p such

that the coordinate curves are the integral curves of X and their orthogonal trajectories. Namely, we may assume that X = X I & and g12 = (&, &) = 0. Because the flow X t of X for every t is an isometry, the vector field X should satisfy the Killing equation Xi, j + Xj, i = 0; i, j = 1,2 , or equivalently,

(Indiana)

218

Taking i = 1, j = 2, we obtain 911% = 0, i.e., X1 = X1(xl). Now, make the following coordinate transformation

If we still adopt the original notations, then the vector field X = &. Hence from the corresponding killing equation, taking i = j = 1 and i = j = 2, we obtain

- = o , -- 8x1 d x i - 0 ,

a922 agll

which means that the metric of M 2 about p can be written as

ds2 = g11(x2)(dx1)2 +- g 2 2 ( ~ ~ ) ( d ~ ~ ) ~ .

Then, using the Liouville formula, we can compute the geodesic curvature of the xl-coordinate curve as follows:

ak where 0 = 0. Therefore, along every integral curve of X, = 0, i.e., Ic , = const.

ii) Analogously, about p , take the integral curves of X and Y as the x1 and x2 coordinate curves, respectively. Then, we have locally X = X I & ,

Y = Y2&. Because their flows are commutative X t o Y, = Y, o X t , that is equivalent to

lay2 d ax1 d [ X , Y ] = x -- - Y 2 - - 8x1 a x 2 dx2 dXl = O ,

we immediately have

= 0. dX1 - = o , - d Y 2

axl ax2 Therefore, we can make a suitable coordinate transformation and then, if adopting the original notations, X = &, Y = G. Again, by the corresponding Killing equations, we can obtain a = 0; i, j , Ic = 1 ,2 , i.e., all gij's are constants. Noting the expressions of X and Y , we obtain

a

( X , X ) = gll = const, ( X , Y ) = g12 = const, (Y, Y ) = 922 = const.

219

3328

Let M be a compact Riemannian manifold without boundary. For any

At any p E M , choose an orthonormal frame field { e l , . - . , e , } around p f E Cm(M), define O f E X ( M ) and Af E C“(M) as follows:

and then define

(Vf)(p) = C ( e i f ) ( e i ) and (Af>(p> = - C [ e i ( e i f ) - ( v e i e i ) f I - i i

Verify first that V f and A f are well defined (i.e., they do not depend on the choice of orthonormal frame) and then show that

(Indiana) Solution.

may suppose that Let { e ; , - e . , e i } be another orthonormal frame field around p . Then we

e i = C a ! e r i = l , . . . 1n

j

for suitable functions a!, i , j = 1,. . . ,n. Noting

(ei , e j ) = (e: , e3) = S i j ,

we have

j

Using these equalities and the properties of Riemannian connection, we can easily obtain

i i

C [ e i ( e i f ) - ( v e , e i ) f l = C [ e r ( e f f ) - ( v e ; e : ) . f ] . i i

Hence Of and Af are all well defined.

have div( f V f ) = llVf112 - f A f. Then Green’s theorem implies Using the definition of the Laplacian here, through direct claculation, we

220

3320

For what a , b 2 0, is Sa,b a manifold? Explain.

Solution. Denote a = xf + x i and ,f3 = x i + x z . Then C Y ~ = a and CY + p = b

means that a and ,f3 are two roots of the equation X 2 - bX + a = 0. Therefore, b2 - 4a 2 0 is the prerequisite condition.

1. If a = b = 0 , then 21 = 2 2 = 2 3 = 2 4 = 0. Thus S0,o is a 0-dimensional manifold.

2. If a = 0, b # 0, then

(Indiana)

= {Z E B4 1 x1 = 2 2 = 0 , x : + 2 : = b or 2; i -2: = b , x 3 = 2 4 = 0).

Using the theorem of closed regular submanifolds proved by rank theorem, we can easily show that &,b is a 1-dimensional submanifold of R 4 .

3 . If a # 0, b # 0, then when b2 - 4a > 0

and when b2 - 4a = 0

b S a , b = (2 E E4 I + 2; = 2; + 2; = 2}.

Analogously, we can show that they are all 2-dimensional submanifolds of R4.

3330

Let F : M -+ N be a C" map between two C" manifolds. Assume that

(i) Show by an example that d F ( X ) may not be a vector field on N . F is onto. Let X be a smooth vector field on M .

221

(ii) Suppose Y = d F ( X ) is a smooth vector field on N . Show that F takes

(iii) Suppose XI, Y1, and X z , Yz are related as X and Y in (ii) above. Show integral curves of X into integral curves of Y.

that d F ( [ X l , X Z l ) = [Yl,YZ]. (Indiana)

Solution. (i) Let M = {(x,y) : z,y E IR}, N = {x : x E E}, x = (xz + y 2 ) & ,

and F : M + N be defined by F(z,y) = x. Then, if y1 # yz, we have d F ( X ( x , y l ) ) # d F ( X ( z , yz)). Therefore, as a vector field, d F ( X ) is not well defined in every point of N .

(ii) Let a(.) be an arbitrary integral curve of X . Then, along a , we have da ($) = X . Hence,

d ( F o a ) (i) = ( d F o da) (g) = d F ( X ) = Y,

which means that F takes integral curves of X into integral curves of Y .

desired equality. (iii) Using local coordinates, by direct computation, one will obtain the

3331

Let M" be a Riemannian manifold. Show that whenever f : M + lR is a smooth function, there is a unique vector field V f (called the gradient of f on M ) such that

whenever y is a smooth curve in M with y(t) = p .

Solution. Let (xl,. - - , xn) be the local coordinates about p and set p : (xk,. . , x2;).

If y ( t ) is the i-th coordinate curve that passes p , i.e., y(t) has the following expression

(Indiana)

21 1 t , XJ 1 x 3 0 (.i # 4, then, by

222

we have the expression

where gij's are the components of the matrix [gijl-' = [(&, It is easy to verify that the above expression of V f (p) is independent of the choice of local coordinates.

3332

The space ELnxn of n x n real matrices forms an n2 dimensional Euclidean space, in which the dot product between A = [a;j] and B = [bij] is given by

n n

(A , B ) := a,j bi j . i= l j = 1

Let Sn-' be the unit sphere in IR", and define u : Snw1 -+ Enxn as the map sending x = (21,. a , 2,) in S"-l to the symmetric matrix ~ ( x ) = I [ z i z j ] .

i) Show that u maps S"-' into the sphere of radius 5 centered at the origin in R" n.

ii) Prove that cr is a local isometry (i.e., the pull-back via g of the dotpro- duct metric defined above on RnX" is the standard one on S"-').

Solution.

Jz

(Indiana)

n

i=1 i) Let x = (zl,..-,zn) E S"-l. Then C z; = 1. Hence

1 " 1 (u(x),u(x)) = - c z;z; = 5 ,

i , j=1

which means that u(x) is on the sphere of radius 2 centered at the origin in

ii) Let i : S"-I -+ Rn be the standard inclusion map, and r : an -+ RnX"

4 5 n n x n

be defined by (21,. . . ,z,) -+ [yij] = -&[zizj]. Suppose that

223

belong to Tx(Sn-') . Then we have

n n

i= 1 i = l

Therefore

Hence we have . n n 1

(dn(v),dn(w)) = - c ( d X j + V j X i ) ( W i X j + W j X i ) = c viwi = (v, w). i , j = 1 i= l

3333

Let M be the Riemannian manifold obtained by equipping B" with a metric conformal to its usual one; i.e., a metric of the form [sij] = e 2 f [ S i j ] ,

where f : IR" ---f IR is a smooth function, and Si j is the Kronecker delta. Let e i = =& denote the standard coordinate basis vector fields.

i) Show that for arbitrary indices i, j , k E { 1 , 2 , , n}

ii) Show that when n = 2, the sectional curvature of M along an e 1 , e 2

plane at p is given by

(Indiana)

224

Solution. i) Set

From g i j = e 2 f & it follows that

ii) The sectional curvature of M along an e l , e2 plane at p is just the Gauss curvature of M at p . Noting that E = G = e 2 f , F = 0, from the Gauss equation we obtain

3334

Let X , Y be complete vector fields on a manifold M and let X , , Yt be the

i) Show that X , o r, = yt o X, for all s, t E IR implies [ X , Y ] = 0. ii) Prove the converse to i).

flows induced by them.

(Indiana) Solution.

Y is F-invariant, i.e., F,Y = Y , if and only if F o yt = yt o F , W E R.

phism for each s, then Y is X,-invariant. Thus

Observe that for a diffeomorphism F : M -+ M , the complete vector field

i) Suppose that X, o yt = yt o X , for all s, t E R. Since X , is a diffeomor-

1 [ X , Y ] = L X Y = lim -[Y - x,,Y] = 0.

3 4 0 s

ii) Now suppose that [ X , Y ] = 0. Then

225

Hence for each p E M and any f E C"(p), we have

for all s E R. Therefore,

( X s * Y ) p f = ( X o * Y ) p f = Y p f .

Since f is arbitrary, we obtain X , , Y = Y , namely, Y is X,-invariant. Hence X , ox = yt OX, for all s , t E R.

3335

If 4 : G1 -+ G2 is a Lie group homomorhism, show that for all w E LG1, we have

4(exp(u)) = exp(+*v).

(Indiana) Solution.

Note that exptw, t E lR' is a 1-parameter subgroup of G1 generated by v E LG1. Since 4 is a Lie group homomorphism, then 4(exptv) is also a 1-parameter subgroup of G2 generated by a suitable w E LG2, namely, 4(exp tv) = exp tw.

. , yn) be local coordinates of G1 and G2 about the identities el and e2, respectively. Locally, 4 can be expressed by

Let (z l , . . . , z") and (y',

and exp tv and exp tw can be denoted by

respectively. Assuming that

n " a a w e l = c v i - , we2 = E w e - aye '

e= 1 axi

i = l

226

we have

t = O t=o Therefore,

= 2 (Z) - = w e , . a (I= 1 t = O aY"

Furthermore, by left translation, we have d $ ( v ) = w, i.e., d*w = w. Hence 4(exptw) = expt4,w. Taking t = 1 completes the proof.

3336

Consider the linearly independent vector fields r and v on U := B4\0 whose values a t x = ( 2 1 , 2 2 , 2 3 , 2 4 ) E B4 are given by

rx : = ( 2 1 , 2 2 , 2 3 , 2 4 ) ,

v , : = ( - 2 2 , 2 1 , - 2 4 , 2 3 )

a) Is the rank-2 distribution defined by these two vector fields in U com-

b) Is the rank-2 distribution orthogonal to these two vector fields completely pletely integrable?

integrable?

Solution. a) Direct calculation gives [r,v] = 0. Thus the Frobenius theorem guar-

antees that the rank-2 distribution defined by r and v in U is completely integrable.

b) Construct the following linear algebraic equations about y1, y2, y3 and Y4

(Indiana)

Z l Y l + 2 2 y 2 + 23Y3 + 24Y4 = 0, I - 2 2 y 1 + ZlYZ - 2 4 y 3 + 23Y4 = 0, which are equivalent to

227

Because ( 2 1 , ~ 2 , 2 3 , 2 4 ) E B4\0 = U, without loss of generality, we may assume 2 1 # 0. Therefore, we obtain

2 1 2 3 + 2 2 2 4

2 2 2 3 - 2 1 2 4

2 1 2 4 - 2 2 2 3

2 2 2 4 + 2 1 2 3

Setting

and

respectively, we obtain the following two linearly independent vector fields CY

and p which define a rank-2 distribution orthogonal to the above one and whose values at are given by

= ( 2 1 2 3 + 2 2 2 4 , 2 2 2 3 - 2 2 1 2 4 , -(.s+.:), o), p = ( 2 1 2 2 - 2 2 2 3 , 2 2 2 4 + 2 1 2 3 , 0, -(.: -k 2;)).

By a long but straightforward calculation, we have [ c Y , ~ ] = -2(2: + zi)v which does not belong to the distribution defined by CY and p, Thus the F’robenius theorem tells us that this distribution is not completely integrable.

3337

Let G be a Riemannian manifold with a global frame-field {ei}y=l.

a) Show that any connection on G is competely determined by its effect on the frame field, i.e., by the vector fields Veiej, i,j = 1,

b) Show that when G is a Lie group with a bi-invariant metric ( , ), and the frame-field is left-invariant, we characterize the Levi-Civita connection on

Veiej = - [ e i , e j ]

, n.

(G, ( 7 )) by setting, 1 2

for all i, j = 1, - . . , n. (Indiana)

Solution. a) For arbitrary smooth vector fields X and Y on G, we have

n n

228

Then, motivated by the properties of connection, we can well define

n

x ' e ; ( g ) - e j + C zigveiej . i , j=l

Thus defined operator V certainly satisfies all properties of a linear connection on G.

b) For every left invariant vector field X on G, let g ( t ) denote the unique 1-parameter subgroup of G such that g(0) = e , 9 = X , , where e is the unit element of G. Noting that g ( t ) is a geodesic of G and Xg(t l = y, we have obviously

vx,x = - t = O

Besides, because the metric of G is bi-invariant, we know that V x X = 0 is valid everywhere. Especially, if X = ei + e j , then V e i + e j ( e i + e j ) = 0 implies that

Veiej + Veje, = 0.

On the other hand, the Levi-Civita connection V satisfies

V,,ej - Veje i = [e i ,e j ] .

Thus, we obtain 1 2 Veiej = - [e i ,e j ] .

Part IV

Real Analysis

231

SECTION 1 MEASURABLITY AND MEASURE

4101

Let S c [0,1] be the set defined by the property that z E S if and only if in the decimal representation of z the first appearance of the digit 2 precedes the first appearance of the digit 3. Prove that S is Lebesgue measurable and find its measure.


For any z E [0,1) there is a unique sequence {pn(z))r'l of integers satis-

(1) 0 5 p n ( z ) 5 9, (2) Vn, 3rn 2 n : p , ( z ) 5 8 and (3) z = p*.

fying the following properties

n = l Then

s = (1) u {. E LO, 1) I h P n ( z ) # 2 and Pn(.) # 3) U{z E [0 ,1) I 3n,pn(z ) = 2 , V i < n,p i (z ) # 2 and pi(z) # 3).

Let A = {z E [O, 1) I V n , p n ( z ) # 2 and p n ( z ) # 3)

and

B = {z E [ O , l ) 1 3n,pn(z) = 2,V i < n,pi(z) # 2 and pi(.) # 3).

Then 00

kn-1 2 k l k - 1 4 10-1 10n 10 + -, - + .. . +- lo*-1 + 1uR)

n=lki#2,3

and

kn-1 2 kl kn-1 3 10n-1 10" l o

+- B = C u [$+.- .+- + -, - + ... n = l kj#2,3

It follows that both A and B are measurable and therefore is S. Since c y ) n

232

and

we have 1 2

m(S) = m(A) + m(B) = -.

4102

Define S1 = IR/Z endowed with the natural Lebesgue measure. Consider on S1 the equivalence relation: [z] - [y] # z - y E &, for z,y E IR and [ ] denoting the class in S1. For each ( E S1/ -, choose a representative f E S1, and let E c S1 be the set of these points, i.e.,

E = { F I E <,V[ E S'/ -}.

Show that E is not measurable.

Solution. S1 is an abelian group under the binary operation ([z], [y]) H [z + y ] and

the inverse operation. [z] +-+ [-.I. For any [z] E S1 there is by the construction of E a unique element [y] E E

such that z - y E &. Then [z] = [y] + [r], where T E [0,1) n& is such that r G z - y modZ. So we conclude that S1 =

( StanfoTd)

U ( E + [T I ) . T €[O,l)W

If T , s E [0,1) n& are such that ( E + [T I ) n ( E + [ s ] ) # 0 there are [z], [y] E E such that [z] + [T] = [y] + [ s ] . It follows that z y mod& and therefore [z] = [y] by the construction of E. Thus [r] = [ s ] , which then implies that T = s since -1 < r - s < 1.

It follows that { E + [TI 1 T E [0, 1) n&} is a partition of S1. Let m denote the natural Lebesgue measure on S1 such that m(S1) = 1.

If E is measurable then

m(S1) = C m ( E + [ r ] ) = C m(E) . T € [O, 1) w T €P,1)W

If m ( E ) = 0 then m(S1) = 0, a contradiction; Otherwise m ( S 1 ) = 00, a contradiction. too.

233

4103

Let X be a set and 'D c ' P ( X ) , 'D closed under finite intersection. Denote by R the ring generated by D. F'uthermore, let a be the smallest system, D c a c P ( X ) such that R is closed under the following operations:

(i) finite disjoint unions. (ii) differences A\B, B c A.

Prove that a = R.

Solution. (Iowa)

Obviously, x C R. We have

since the former is a ring containing D. Also we have

since by the equality (1) the former is a ring containing 2). By equality (2), R = R since for any A E R, A = A n A.

4104

Let p* be the Lebesgue outer measure on IR and A , B subsets of IR such that

inf{ 12 - yI I z E A, y E B } > 0.

Prove or disprove that

p * ( A U B ) = p* (A) + p*(B) .

Solution. We will show the equality. Let T = d(A, B ) > 0. Let

U = {Z E I d(z ,A) < T) 2

(Iowa)

234

Then U and V axe disjoint open sets containing A and B respectively. We have

p * ( A u B ) =

= inf{p(W) I W o p e n , A U B C W C U U V ) =

E inf(p(W1) 1 ~1 open, A 5 W I c U }

= P*(A) + P * ( B ) ,

inf{p(W) I W open,A U B C: W }

inf(p(W1) + p ( W 2 ) I Wi open, A C WI C U , B C W2 2 V }

+inf{p(W2) I W2 open, B C W2 C V}

where the equality (1) follows from the following equality

i n f ( X + Y ) = i n f X + i n f Y , X , Y GR.

4105

a) Consider a measurable space (X, p ) with a finite, positive, finitely additive measure p. Finite additivity means that whenever { B i } is a finite collection of mutually disjoint measurable sets, then p(UBi) = C p(Bi) . Prove that p is countably additive if and only if it satisfies the following condition:

If A, is a decreasing sequence of sets with empty intersection then

lim p(An) = 0. T - 0 0

b) Now that suppose X is a locally compact Hausdorff space, that B is the Bore1 a-algebra, and that p is a finite, positive, finitely additive measure on B. Suppose moreover that p is regular, that is for each B E B,

p ( B ) = sup{p(K) I K 5 B and K is compact}.

Prove that p is countably additive.

Solution. (Iowa)

a) The sufficiency. Let {B,} be countably many measurable sets which are 00 m

mutually disjoint. Let A, = U Bi. Then n An = 0. We have i = n + l n = l

/ n \ o o

235

Therefore ,Y is a measure. The necessity is obvious.

of measurable sets with empty intersection such that b) If /I is not countably additive, by a) there is a decreasing sequence {A,}

lim p(A,) = inf p ( A n ) > 0. ,--too n

For each n there is a compact Kn contained in A, such that

which implies that

and therefore

i=l

n

i=l Thus { Ki I n E N} is a decreasing sequence of nonempty compact subsets

in the compact space K1. So n K , # 0 , which contradicts the fact that 03

n=l

n = l

4106

For f : [0,1] -+ B, let E C {z I f'(z) exists}. If m ( E ) = 0, show that

( Indiana-Purdue) m(f(E)) = 0.

Solution. Denote

F = {. 12 E E , lf'(4I < MI,

where M is any positive number. It suffices to prove m ( f ( F ) ) = 0. Set

1 n Fn = I z E F, If(y) - f(z>I I MIY - 21 if I Y - 21 < -1.

236

Then F1 C F2 C * . . , m(Fn) = 0,

and f(F) c uf(Fn) = lim f(Fn).

n-cu

For any E > 0, take a sequence { I n $ } of open intervals such that

For 2, y E Fn f l In$, we have

Therefore

5 m * ( f ( F n f' I n , k ) ) k

5 M m ( I n , k ) < ME-

Thus m*(f(Fn)) = 0, which implies that m ( f ( F ) ) = 0.

4107

Suppose A c IR is Lebesgue measurable and assume that

for any a, b E IR, a < b. Prove that m(A) = 0.

Solution.

open subset U in (n, n + 1) such that

(Iowa)

If m(A) # 0 there is an n such that m(A n (n,n + 1)) # 0. There is an

A n (n, n + 1) C U C (n, n + 1)

237

where E < m(A n (n, n + 1)). There are at most countably many disjoint intervals (u j , bj)’s such that

Then An(n ,n+ 1) = UAn(a j ,b j ) .

j

We have

which deduces that m(A n (n, n + 1)) < E ,

a contradiction.

4108

Choose 0 < X < 1 and construct the Cantor set Kx as follows: Remove from [0,1] its middle part of length A; we are left with two intervals 11 and 12. Remove from each of them their middle parts of lengths X l l i j l , etc. and keep doing this ad infinitum. We are left with the set Kx. Prove that the set Kx has Lebesgue measure zero.

Solution.

step is X ( l -

total length of intervals removed in the k + 1-th step is

(Stanford)

Claim. For any n E W , the total length of intervals removed in the n-th

The claim holds for n = 1. Assume that it holds for any n 5 k . Then the

k

i = l

238

By induction the claim holds for any n E JV. It follows that the Lebesgue measure of Kx is

1 - c X ( 1 - A y - 1 = 0. n = l

4109

Suppose p is a positive Borel measure on lR such that (i) p([O,l]) = 1, and (ii) p ( E ) = p ( z + E) for any Borel set E of 2R and

every z E B. Does this imply that p is the Lebesgue measure? Justify your answer.

Solution. (Iowa)

Yes, p is the Lebesgue measure. For any z E IR define

Then g : lR -+ lR is nondecreasing and right-continuous. Moreover, for any z , y E IR with z < y, p (z ,y ] = g(y) - g(z). It follows that the measure p is induced by g.

For any z , y E B, from either

A.7 z + Y1 = 4 0 , Y1, Y L 0

or p ( z + Y, .I = P(Y, 01, Y < 0

we have g ( z + Y) = g(z) + d Y ) .

g(1) = p((0,11) = P([O, 11) - P ( { W

From the right-continuity of g, we conclude that g(z) = zg(1). However,

1 n+oo n = 1 - lim p((--,O])

1 n+oo n

= 1 - lim (g(O) - g(--)) = 1.

It follows that g(z) 2 z and therefore p is the Lebesgue measure.

239

4110

Let U be a a-algebra of subsets of a set X and p n : U -+ lR be signed measures such that p ( E ) = lim p n ( E ) exists for every E E U. Prove that ,u is a signed measure.

Solution.

ndoo

( Iowa)

Let, for each E E U,

Then ,Z is a finite measure and pn’s are absolutely continuous with respect to ,G. Given any mutually disjoint measurable sets En’s, by the Vitali-Hahn-Saks

Then

k=l

which shows that p is a generalized measure and therefore a signed measure.

4111

Let X be Lebesgue measure on R. Show that for any Lebesgue measurable set E c 1R with X(E) = 1, there is a Lebesgue measurable set A c E with X(A) = !j.

( Iowa)

240

Solution. Define the function f : IR -+ [0, 11 by

f(2) = X(E n (-m,z]),

If(.) - f ( Y ) I 5 12 - YI, Z,Y E E.

2 E E.

It is continuous by the following inequality

Since

f(z0) = f. Put A = E n (-co,zo], which is required.

lim f(z) = 0 and lim f(z) = 1, there is a point 20 E E such that x + - w x+ w

4112

Let p be a complex Bore1 measure on [0, co). Show that if rw

for all n = 0,1 ,2 , . . ., then ,u = 0.

Solution. Let

S = span{ePa5 I n E lN u (0))

(Iowa)

then S is a self-adjoint subalgebra of C( [0, co]) separating the points of [0, co]. It follows from the Stone-Weierstrass Theorem, S is dense in C([O, m]) under the supremum norm topology. Therefore for any f E C([O, oo]),

which then implies that p = 0.

4113

Let A c [O, 11 be a measurable set of positive measure. Show that there

( Indiana-Purdue) exist two points z/ # z” in A with 2’ - z” rational.

Solution. Denote all the rational numbers in [-1,1] by rl, 7 3 , . . . , T,, . . .. Denote

A, = {z + T, 12 E A } .

Then m(A,) = m(A) > 0. A, c [-I, 21. Thus

24 1

00

U An c [-1721. n=l

Suppose that A, n Am = 0 if n # m. Then 00

C m(An) I m([-1,21) = 3, n=l

which contradicts m ( A ) > 0. Therefore there must be some m, n such that A n n Am # 0. Take 2 E An n Am. Then we can find z', 2'' E A such that

I1 z = 2' + r, = 2 + T,.

Thus 'I'm - T,,. - =

4114

2, = 0 or 1 . Prove that E is Lebesgue measurable with

(Iowa) Lebesgue measure 0. Also show that E + E = [0,1].

Solution. Obviously, E c [0,1). We will show next that

+- , -+ . - -+ 3n-1 xn-l + 7 2,-1 2 2 1

3 3-1 3" 3 [O, 1)\E = u { [" + * * - + -

00

Indeed, for any 2 E [O,l)\E, 2 = ?$ (where 0 I z, I 2 and for any n

there is an m 2 n such that 2 , I l), there is at least one n such that 2, = 2. Let n = min{k I 21, = 2) then 0 5 zl,...,xn-l I 1 and therefore

n = l

2,-1 2 21 [? 3"-1 3"' 3 a : € - + + * a + - + - -+...

The converse inclusion is obvious. By equality ( l ) , E is measurable and since

242

m(E) = 0. Since E C [0, i], E + E [0, 13. For any z E [0,1) say z = C F (where 0 5 z, I 2, and for any n, there is an m 2 n such that 2, # 2). Define

03

n= 1

and

n E N . z, = 2,

X" 0 0 , 00

n= 1 ,=l Then z' = C 3 and 2'' = C $ belong to E so that ~'$2'' = z. Obvior;sly,


4115

Let f : It2 -+ IR+ be measurable, and let E > 0. Show that there exists g : I22 -+ El+ measurable such that (i) [ I f - gllm 5 E (ii) for every T E B,

(Indiana-Purdue)

Take {r,} such that 0 < r1 < r2 < . . . < T, < . . ., limr, = +m, T,+~ -

I{. I g ( 4 = .}I = 0.

Solution.

r , < E , for all n. Set fl(z) = f(z) + arcctgz, Denote

En = {z I rn-1 < fi(z) I ~n}. Set g1 = T n X E , . Then

g1(z) 2 acc tgz , llfl - glllcx, L E .

Set g(z) = gl(z) - arcctgz. We have g 2 0 and [ I f - gllco = For every r E B,

- gl)loo 5 E .

{zIg(z) = r } = =

{z 1 gl(z) - arcctgz = T }

U{z I z E Enlarcctgz = r, - TI .

n

Since \ ( r c \ arcctgrc = T, - T } \ = 0, \ {z g(z) = T } \ = 0.

243

4116

Let f be the function on [0,1] defined as follows f(z) = 0 if z is a point if 2 is in one of the complementary on the Cantor ternary set and f (z ) =

intervals of length 3 - P .

a) Prove that f is measurable.

b) Evaluate 1 f(z)dz. 1

(Stanford) Solution.

Let K denote the Cantor ternary set. Then

[O, 1]\K = u u (0 . a 1 . . . Up-11,O . a 1 a 2 . . * a p - 1 2 ) . p 2 1 ai=0,2

By the definition, we have

which then is measurable, and

4117

Let E be a Lebesgue measurable subset of lR with m ( E ) < 03 and let

f ( ~ ) = m ( ( ~ + 2 ) n E ) .

Show that (a) f ( z ) is a continuous function on R. (b) lim f(z) = 0.

z++w (Illinois)

Solution. We will show first that for any Lebesgue measurable set E ,

lim m((E + h) n E ) = m ( E ) . h-0

244 n

r = l If E is of the form .G (a;)oi)) where ( q , @ i ) ’ s are finitely many mutually

disjoint open intervals of finite length, then

n

5 h m ( E n ( E + h ) )

<_ h+O

lim m ( E n ( E + h ) ) 5 m(E) . h+O

so lirn m ( ( E + h) n E ) = m ( E ) .

If E is a compact set, then there is, for any E > 0, an open set E’ of the above form such that E c E’ and m(E’\E) < E . Let F = E’\E one has

h-O

m( E ) < m( El) = lim m( (E’ + h) n E’) h-iO

= l i m m ( ( ( E + h ) n E ) u ( ( E + h ) n F ) U ( ( F + h ) n E ) G ( ( F + h ) n F ) )

5 h-0

h m ( ( E + h) n E ) + 3.5 5 G m ( ( E + h) n E ) + 3~ 5 m ( E ) 3 . 3 ~ . h+O h-+O

It follows that m ( E ) = lim m ( ( E + h) n E ) .

In general, there is an increasing sequence { E n } of compact sets such that En C E and

lirn m(E,) = m ( E ) .

h-0

n+m

Then

m ( E ) = lim m ( E n ) = lim lirn m((En + h) n En) n+m n+m h-10

5 lim lim m ( ( E + h) n E ) = lim m ( ( E + h) n E ) . n-+m h+O h+O

245

so lim m((E + h) n E ) = m(E) . h-0

(a) For any y, z E 2R

If(Y) - f(.)I = Im(((E + Y)\(E + n E ) - m ( ( E + z)\(E + Y))l 5 m( (E + Y)\(E +

m(E + (Y - z)\E) + m(E + (z - y) \E) = m ( E ) - m(E + (Y - z)) n E ) + m ( ~ )

+ M ( ( E + z)\(E + Y)) =

-m((E + (z - y)) n E ) -+ 0, as y -+ z.

(b) If E is compact, then there is an T > 0 such that for any z > T ,

In general there is an increasing sequence { E n } of compact sets such that (z + E ) n E = 0. The claim follows.

En E and lim m ( E n ) = m(E) . Then

lim m ( ( E + z) n E )

lim m ( ( E + z) n E ) - m((En + Z) n En) + m((En + Z) n En)

lim lim m ( ( ( E + z) n E)\((En + z) n En)) + m((En + z) n En)

lim ( m ( ( E + .)\(En + z)) + m(E\En)) = 0.

n-cc

X-++OO

z-r+cc =

=

5 n-cc x++m

n-rn

4118

Let A c lR be a set of positive Lebesgue measure. Prove that

cp(z) = J X & ) X A ( t ) d t

is continuous a t z = 1. Use this result to prove that there exists an E > 0 such that for any m E 2R with Im - 11 < E , the line y = mz has a non-void intersection with A x A.

(Iowa) Solution.

If B is a bounded open set, say u (a*, b i ) , where 2 5 m 5 03 and (aj , bj)’s

are mutually disjoint open intervals, then for any n < m and any z > 0 we have

i<m

n

U ( z a i , z b i ) n (a i , b i ) c ZB nB. i = l

246

We have n

p ( B ) = sup C(bi - . a ) n < m . a = 1

n

5 ~ P ( z B ~ B ) 2-1

5 X’l lim ~ ( Z B n B ) 5 ~ L ( B ) , (1)

where p is the Lebesgue measure. If K is a compact set, there is a decreasing

sequence {B,} of bounded open sets such that K = n B,. Since 00

n = l

p(zB, n Bn) - p ( z K n K ) I P(Z(B,\K)) + P(Bn\K)

we have by (1)

In general, there is an increasing sequence {K,} of compact sets such that 00 aJ

U K , C A and p(A\ u K,) = 0. By (2) we have n = l n= 1

Since

247

If there is no E > 0 with the property mentioned in the question, then for any n E mT there is an m, E B2 such that Imn - 11 < and the line y = mnx has a void intersection with A x A. However, this implies that m,(A n A ) = 0, and therefore

p ( A ) = lim p ( m n ( A n A ) ) = 0, n-03

a contradiction.

4119

Let p be a countably additive measure on a set S with p(S) < $00 that is without atoms, i.e., if A is a measurable set with p ( A ) > 0, then there is a measurable set B c A such that 0 < p ( B ) < p ( A ) . Prove that the range of p is the closed interval [0, p(S)].


If there is a t o E (O,p(S)) not in the range of p. Let

P = { A I A measurable and p ( A ) < t o } / -,

where - is an equivalence relation: A - B if p(A\B) = p(B\A) = 0. Then P is a partially ordered set: [-41 5 [B] if p(A\B) = 0. Given a totally ordered set Q of P , let ,B = sup{p(A) I [A] E Q } . Then P 5 t o and there is an increasing

sequence { [ A , ] } of Q with { p ( A , ) } increasing to p. Let A = u An, then 03

n = l

= lim p(A , ) = P 5 t o . ,403

SO p ( A ) = /3 < to and therefore [A] E 7'. For any [B] E Q, if [B] 5 [An] for some n, then [B] 5 [A]; otherwise, [B] = [A] since

03

P(A\B) 5 C p(An\B) = 0 n = l

248

It follows that [A] is an upper bound of Q in P. There is by Zorn’s Lemma a maximal element [Ao] of P. It follows that p ( B ) > t o - p(A0) whenever B C S\Ao and p ( B ) > 0. (e.g. p(S\A) > t o - p(A0) ) . Let

R = { B I B c S\Ao,P(B) > O}/ -, where - is defined as above. Equip 72 with the partial order 5 as above. Given a totally ordered set S of R. Let

a = inf{p(B) I [B] E S},

then cr 2 t o - p ( A ) and there is a decreasing sequence { [Eln]} of S with {p(B, )} decreasing to a. Let

n = l

then

So p ( B ) = a > 0. Therefore p ( B ) > t o - p(A0) . [B] E R. For any [Cl E S, if [Bn] I [C] for some n, then [B] 5 [C]; otherwise [B] = [C] since

n=l

and ~ ( B \ C ) = p ( B ) - p ( B n C ) = - a = 0.

It follows that [B] is a lower bound of S in R. There is by Zorn’s Lemma a minimal element [Bo] of R. However, by the assumption that there is a subset CO of Bo with 0 < ~ ( C O ) < ~(Bo), [CO] 5 [Bo] and [CO] # [Bo], a contradiction.

4120

Let X be a compact Hausdorff space. Consider the c-algebra 23 generated by the compact Ga sets. Show that any positive measure p on B which is finite on compact sets is automatically regular.

(Iowa)

249

Solution.

a Ga set. Also recall that E is outer regular if Recall that sets in 23 are called Baire sets and each compact Baire set K is

p ( E ) = inf{p(V) 1 E C V,V open and V E 23)

and E is inner regular if

p ( E ) = sup{p(K) 1 K C E , K compact and K E B } .

Let K and 0 denote the classes of compact sets and open sets in 23, respectively. Let

? a = { ,U (Ki\Li) I Ki, Li E K}, ¶=1

where the symbol 0 means disjoint union. Then 7E is a ring such that 23 =

By definition, each set in K is outer regular. Let us show each set U in 0 is inner regular. For any c > 0, there is a V in 0 such that X\U C V and p ( V ) < p(X\V) + E . Then X\V 2 U and p ( U ) < p(X\V) + E .

Let us preceed in five steps to show the regularity of p. Step 1. For any pair K , L in K, K\L is regular. For any E > 0 there are

a B E 0 and an M E X such that K C B and p(B\K) < E and such that M C B\L and p(B\L) < p ( M ) + E . Then we have

V).

K\L c B\L, P(B\L) < P(K\L) + E ;

K n M C K\L, p(K\L) 0, there are B1,. . . , B, E 0 such that Ei C Bi with E

p(B,\E;) < ;, i = 1, .. . , 7 ~ .

Consequently, 6 Ei C (J Bi i = l i = l

250

and

n

i= 1 which shows the outer regularity of u Ei. The inner regularity follows from

the following inequalities

Step 3. If {En I n E IN} is an increasing sequence of regular sets then 03

(J En is regular. n = l

00

Let E = u En. Obviously n = l

p ( E ) I inf{p(V) I E G V E O}.

For any E > 0 there are V1,. . , V,, . . . E 0 such that En C Vn and p(Vn) < 03

p ( E n ) + &. Let V = U Vn E 0, then n= 1

which shows the outer regularity of E , while the inner regularity follows from the following inequalities

co Step 4. If { E n } is a decreasing sequence of regular sets, then n En is

n=l regular.

251

00

Let E = n En. For any E > 0, n E lN, there is a K , E ?E such that

n= 1

and

which shows the inner regularity of E. The outer regularity follows from the following inequalities,

inf{p(V) I E V E 0} 5 inf inf{p(V) I En 5 V E 0)

= inf,u(En) = p ( E n

Step 5. Let S = { E E 23 1 E is regular }.

By steps 1 and 2, S contains a. By steps 3 and 4, S is follows that S = B.

t monotone class. It

4121

Suppose that p is a nonnegative Bore1 measure on En. Let

Assume f ( ~ ) is finite for all T > 0 and assume

Prove that p is identically 0.

...

(Indian u)

> n>

252

Since C(z, T ) c B ( z , &?Fir),

Given a compact set K of IR", let s > 0 be such that K c C(0, s). For any E > 0 there exists an T > 0 such that g ( T ) < ( ~ T ) ~ E . There exist finitely many zl,. . . , zk E IRn such that

k

K c u C(z"7-) c C(0,2s) i = l

and C(z*, T) 'S are mutually disjoint. Then

k k

i= 1 i=l k

= ZA(C(z$ T ) ) & 5 X(C(0, as))&. i= 1

Letting E -+ 0, we get that p ( K ) = 0 and therefore p is identically 0.

4122

Let 1.1 be a finite Bore1 measure defined on lRn, and define a function f on En by f (z ) = p ( B ( z , 1)) where B ( z , 1) denotes the open ball centered at z of radius 1. Prove that f attains its minimum on each compact set of IR".

Solution.

exists an sequence { z k } of K such that

(Indiana)

Let K be any nonempty compact set of 1R" and let a = in f f (K) . There

lim f ( z k ) = a . k + o o

Assume without loss of generality that % k + z in K . It is easy to show that

B ( z , l ) B ( X k 3 1 ) - k + m

Hence

253

which shows that P ( B ( X , 1)) = Q.

Hence a is finite and f attains its minimum on K .

4123

Let E be the set of all numbers in [0,1] which can be written in a decimal expansion with no sevens appearing. Thus,

= 0.2800 . . . E E. 28 = 0.2699.. . - 1 27

- 1 0 . 3 3 3 . . . - 3 ' 100 ' 100

(i) Compute the Lebesgue measure of E. (ii) Determine whether E is a Borel set.

(Indian a) Solution.

For any z E [0, l]\E, write z = 0.a1a2 . . .a, . .-. Let

n = min{k 2 1 I ak = 7).

If ak = 0 for all k > n then

3: = 0.611a2 * * * Un-17 = O.aia2 * * * ~~-1699. . . E E, a contradiction. It follows that

0.Ulaz...Un-17 < x < O.alaz...~~-lS. and therefore

[O,l]\E U{(O.ala2...an-17,0.UlU2...~~-18) I ~l,...,an-l # 7:

n = 1,2'. . .}. The reverse inclusion is obvious. So E is Borel measurable and

W 1

m ( E ) = 1 - 9"-l x - = 1 - 1 = 0. 10" n=l

4124

Let f : IR -+ R" be a function such that for all z, y E R

If(.) - f(y)ln 5 e'zl+lyI 12 - yI. (*>

(Indiana) Show that if E c IR is a measurable set with m l ( E ) = 0 then m,( f (E) ) = 0.

254

Solution. Assume without loss of generality that E is bounded. Let E 2 (-r,r) .

For any E > 0 there exists an open set U such that E U C ( -T ,T) and rnl(U\E) < E . Write U = U(uilbi) , where (ai ,bi) 's are mutually disjoint.

Then condition (*) implies that i

f ( (a i ,b i ) ) c ~ ( f ( a i ) , (eZrlbi - ail) ') .

It follows that

r n , ( f ( ~ ) ) _< C C,e2'Ibi - ail < CneZr&,

which implies that rn:(f(E)) = 0 and therefore f ( E ) is measurable. Hence

a

mn(f(E)) = 0.

4125

Let f : R" -+ lR be an arbitrary function having the property that for each E > 0, there is an open set U with X(U) < E such that f is continuous on E"\U (in the relative topology). Prove that f is measurable.

Solution.

Let fk = fXEn\u,, then fk is measurable. For any E > 0,

(Indiana)

and f is continuous on R n \ u k . Let u k be an open set such that x ( u k ) <

1 m*({z I l f k - f I(.) 2 E l ) = m*({z E u k I If(x)I 2 E } ) 5 C.

It follows that {fk} converges to f in measure. Since the Lebesgue measure is complete f is measurable.

4126

Let A c R be a Lebesgue measurable set and let r A = { r A I z E A } where T is a real number. Assume that r A = A for every nonzero rational number r. Prove that either A or R\A has Lebesgue measure zero.

(Indiana)

255

Solution. Let lR* = R\{O} and 3 = A\{O}. Then rB = B and T(R*\B) = R*\B

for every nonzero rational number T . If m(lR*\B) = m(R\A) > 0 there exists a compact subset K of E*\B with positive Lebesgue measure. For any compact subset L of B , define function f : lR* + [0, cm) by

dY f(2) = 1,. XK(Y)XL-l(Y-lz)-, 2 E a*. Y

Then f is continuous and for any nonzero rational number T ,

Hence f( z) = 0 for any z E R* . Since

we conclude that m ( L - l ) = 0 and therefore m ( L ) = 0. It follows that m ( B ) = 0 and therefore m(A) = 0.

4127

Let p be a c-finite measure on the measure space ( X , m). Prove that there exists a probability measure Y on ( X , m ) ( ~ ( z ) = 1) such that p is absolutely continuous with respect to v, and v is absolutely continuous with respect to p.

Solution. (Indiana)

There exists a sequence {En}r=p=l of mutually disjoint measurable sets of

finite and positive measure such that X = u En. For each measurable set E

let

QJ

n = l

then Y is the desired probability measure.

256

SECTION 2 INTEGRAL

4201

Prove or disprove that the composition of any two Lebesgue integrable functions with compact support f , g : IR -+ IR is still integrable.


It is not true. For example, let

f(.> = X { 0 } ( 4 and d.1 = x{o,l}(x).

Then f and g are integrable functions with compact support. However, since g o f (z) 5 1, the function g o f is not integrable.

4202

Let f E & ( O , 1). Assume that for any x E ( 0 , l ) and every E > 0, there is an open interval J, 2 ( 0 , l ) such that

x EJ,, m(J,) < E , and fdm = 0.

Prove that for every open interval I E ( 0 , l )

fdm = 0 .

(Illinois)

There is a measurable set E of measure zero such that any x E (0, 1)\E is Solution.

a Lebesgue point o f f , i.e.,

257

For any 2 E (0, l)\E and for any n there is an open interval Jn & ( 0 , l ) such that z E J,, m(J,) < and

It follows by equality (1) that f (z ) = 0, i.e., f (z) = 0 a.e.. Therefore

4203

Let (X, M , p ) be a positive measure space with p ( X ) < 03. Show that a measurable function f : X + [0, 03) is integrable (i.e., one has Jx f d p < CO)

if and only if the series 00 c I f (z) 2 n})

n=O converges.

Solution. (Iowa)

Suppose that f is integrable. Then W

258

Conversely,

which shows that f is integrable.

4204

(a) Is there a Borel measure p (positive or complex) on IR with the property that

J , fdp = f (0)

for all continuous f : B + (c of compact support? Justify.

that (b) Is there a Borel measure p (positive or complex) on B with the property

J, f d p = f’(0)

for all continuously differentiable f : R + (c of compact support? Justify.

Solution. (Iowa)

(a) Yes. Let p ( E ) = x ~ ( 0 ) for any Borel set E. (b) No. If there were such a Borel measure, let ‘p 2 0 be a continuously

differentiable function of compact support, taking value one on [-1,1]. Then a contradiction occurs from the following limits

JL% Irn ( p ( t ) e t d t = /, p ( t ) d t > 0

and

4205

Let E be a Banach space, (X,?r , p) a probability space, and f : X -+ E such that g o f is p-integrable for every g E E‘.

259

Define L : E‘ -+ IR, L(g) = g 0 fdp. I

Does L E E”? Justify your answer. (Iowa)

Solution. It is ture that L E E”. Define the linear operator

T : E’ + L1(X), p H p o f .

Assume that pn -+ ‘p in E’ and T(pn) + h in L1(X). I t follows that pn o f converges t o p o f everywhere and to h in measure and therefore h = p o f in L 1 ( X ) . By the closed graph theorem, T is bounded. We have

So L is bounded.

4206

Let ( X , M , p ) be a positive measure space with p(X) < co, and let f and g be real-valued measurable functions with

Show that either (a) f = g a.e., or (b) there exists an E E M such that

(Iowa) Solution.

If (b) does not hold, then for any E E M ,

Since

260

for any E E M ,

J , f d P = J, d P .

For any E > 0 the sets

are measurable. From the equalities

and

we conclude that p ( E + ) = p ( E - ) = 0. I t follows that

Therefore (a) holds.

4207

Let ( X , M , p ) be a positive measure space, and S a closed set ina'. Suppose that

1 --J f @ € S P ( E ) E

whenever E E M and p ( E ) > 0. Show that {z E X I f(z) @ S } has measure 0.

Solution.

many closed balls { z E C I Iz - A, 1 5 E ~ } ' s . If

(Iowa)

Since a' is second countable, one finds that a'\S is the union of countably

261

there is a t least an n such that

P ( { ~ E X I I f ( ~ ) - X n I F E n ) ) # O *

But then & J” f d p belongs to { z E (J‘ I Iz - XnI 5 E ~ } , not to S, where E = {x E X I If(.) - X,I 5 E ~ } ) , a contradiction.

4208

Let f : [0,1] -+ (0, m) and let 0 < a 5 1. Show that

where inf is extended over all measurable E c [0,1] with m ( E ) 2 a.

Solution. (Indiana-Purdue)

0 bviously,

Take an N such that

Suppose that E C [0,1], m ( E ) > a. Denote El = E (f 2 +), EZ = E\E1. Then m(E2) < 5 , so m(E1) > $. Therefore

Thus

262

4209

Let {fn} be a sequence of real-valued functions in L 1 ( B ) and suppose that for some f E L1(IR),

Prove that fn -+ f almost everywhere with respect to Lebesgue measure.

Solution. ( Il Zin o is)

Since

there is, by Levi’s Lemma, a measurable set E of measure zero such that for any t E B\E,

n

Therefore for any t E lR\E,

n

fn(t) = fl(t) -k C ( f k - f k - l ) ( t ) k = 2

converges. It follows that fn + f almost everywhere.

4210

Let /I be a finite measure on lR, and define

Show that f (z ) is finite a.e. with respect to the Lebesgue measure on lR.

Solution. (Indiana)

Let

263

then g E L 1 ( R , d v ) , where

dx dv(z) = -

1 + 2 2 ’

by hbini’s Theorem, the function

is finite a.e. with respect to the measure v. The conclusion follows from that the measure v and the Lebesgue measure are equivalent.

4211

Let ( X , M , p ) be a positive measure space, f n : X + [0, CQ] a sequence of integrable functions, and f : X ---t [0, m] an integrable function. Suppose that fn -+ f a.e. [p], and that

Prove that

(Iowa) Solution.

Assume, without loss of generality, that p is totally a-finite, i.e.,

264

n=l

where Xn’s are mutually disjoint measurable sets of finite and positive measure. Let

Then g is integrable. Let dv(2) = g ( z ) d p ( z ) . Then v is a finite measure on X , equivalent to p. Let

dXn(z ) 1 f n ( z ) d p ( z ) , 71 E m* Then An << v.

For any A E M , {s, f n d p } is bounded and therefore admits a convergent subsequence. If {s, fn ,dp} is any such subsequence, then by Fatou’s Lemma one has

It follows that

r r and therefore

For any E > 0 there is a 6 > 0 such that Xn(E) < E whenever v ( E ) < 6, by the Vitali-Hahn-Saks Theorem. There is by Egroff’s Theorem an E with v ( E ) < 6 such that { $} converges to f 9 uniformly on X\E. Then

265

4212

Let p be a a-finite, positive measure on a u-algebra M in a set X. (a) Show that there exists W E L1(p) which takes its values in the open

interval (0 , l ) . Show also that

r

is a positive, finite measure on M , and that

P(E) = 0 e p ( E ) = 0

for E E M .

respect to M, then (b) Show that if f is a complex function on X which is measurable with

(Iowa) Solution.

03

(a) Let {X,} be a disjoint sequence of M such that X = U X , and n=l

p(X , ) > 0. Let

W=C- O0 1 xx, n=l 2, 1 + p(Xn)

as desired.

then P(E) = 0. Conversely, The set function P is of course a positive, finite measure on M . If p ( E ) = 0

266

(b) If f is a simple function, i.e.,

n

f = C a i x E i ! i = l

where p ( E i ) < +a, i = l , . . - l n , then both sx f d j i and sx fWdp equal to n 2 aiii(Ei).

sequence { fn} of simple functions integrable with respect to p such that

i= 1 In general, if f is integrable with respect to the measure p, there is a

Then fnW’s are integrable with respect to p, and since

lim Jx\jnw-fmWldp= lim ~ f ~ - f m l d j i = ~ , n-+w n-+w

m+oo m+cc

lim fnW = f W , n-cc

we see that fW is integrable with respect to p. We have

and therefore

J , f d F = J X fWdP

Conversely, if fW is integrable with respect to pl assuming without loss of generality that f is positive-valued, there is a sequence of simple functions { f n } such that l f n l 5 f and lim fn = f . Then fnW is integrable with respect to p and hence fn is integrable with respect to p. Since

n-oo

and lim fn = f, f is integrable with respect to p. Accordingly, n+cc

267

4213

Let m denote the Lebesgue measure on [O, 13 and let (fn) be a sequence in

(i) (ii) I fn I Show that

L1(m) and h a non-negative element of L1(m). Suppose that fngdm -+ 0 for each g E C([O, 11) and

h for all n.

J, fndm + 0

for each Bore1 subset A [0,1]. (Iowa)

Solution. For any E > 0, there is a 6 > 0 such that

IE hdm < E

whenever m(E) < 6. For such a 6 there are a compact set K and an open set U such that (1) K A C U and (2) m(U\K) < 6. There is a continuous function g : [0,1] --t R such that (3) 0 5 g 5 1, ( 4 ) g = 1 on K and (5) g = 0 outside U . Then we have

,% IJ,fndml = n-+m lilfnXAdml

It follows that Jiir J, fndm = 0.

4214

Let { fn} be a sequence of non-negative measurable functions in LP( R) for

( Indiana-Pvrdm) some 1 < p < 00. Show that fn 4 f(P) if and only if f: 4 fP(L1).

268

Then fn 5 f and IL - f I L Ifn - fl

which implies that IIE - f ( I p -+ 0. Just as above, we have

Since T:+X = f , ” + f P ,

SO

Therefore

Conversely. Suppose that

J If,” - f p l - + 0.

Then fn 2 f , f > 0.

Since I J u ~ - f P ) / I J tfi - PI -+ 0,

Jfi -+ J f p .

it follows that

For any c > 0, take N1 such that

269

for n > N1 and any measurable set E. Take A, a measurable set, such that m(A) < +m and

J,. f P < & P P

J,. f: < EP

Then

for n > N1. Take S > 0 such that

if m(e) < 6. Thus

Jf.. e < EP

if m(e) < S and n > N1. Denote 7 = &/m(A)f. Take N such that N 2 N1

a n d m ( l f n - f l > q ) < S i f n > N . Wehave

(J, Ifn - f l p )

U P

for any n > N .

4215

Let 1 5 p < 00. All parts refer to Lebesgue measure on R. (a) Give an example where { f n } converges to f pointwise, l l f n l l p 5 M , Vn

and llfn - fllp f , 0.

270

(b) If { f n } converges to f pointwise and l ] f n l l p -+ M < $00, what can you

(c) Show that if { f n } converges to f pointwise and l l fnllp -+ I l f l lP , then

(Indiana-Purdue)

conclude about Ilfllp? Justify this conclusion.

llfn - fllp --+ 0

Solution. (a) Consider L[O, 13. Set fn = n ~ ( ~ , i ) , Then

lim f n ( z ) = f (z ) = 0, z E [O, 11 n-cc

and llfn - fll = llfnll = 1.

(b) We conclude that l l f l l p 5 M . By the Fatou's lemma,

(c) Set gn = 2"(lfn Ip + Ifl") - l f n - f l P .

Then gn 2 0, and lim gn(z) = 2 p + ' l f l p

n-+m

pointwise. Using the Fatou lemma, we have

Therefore I f n - fJPdz = 0.

n-o3

4216

Suppose f n is a sequence of measurable functions on [0,1] with

1' lfn(z)12dz 5 10

and f n --+ 0 a.e. on [0,1]. Prove

1' lfnldz --t 0-

27 1

Hint. Use Egorov and Cauchy-Schwartz. (Stanford)

Solution. For any E > 0 there is by Egorov’s Theorem a measurable set E c [0,1]

such that (1) m([O, 11 I E ) < E and (2) fn converges to 0 uniformly on [O, 1]\E. We have by the Cauchy-Schwartz Inequality

and therefore

Let E + 0, and we have r l

4217

Let (X, M , p) be a probability space (a positive measure space with p ( X ) = 1). Show that i f f is an integrable function with values in [1,00), then

lim P10 (J, f p d p ) ’ = exp ( J , log f dp ) .

(Iowa) Solution.

It suffices to show that

If f = 1 a.e., equality (1) holds; Otherwise, for any 0 < p < 1 and any 2 E X

272

By the Dominated Convergence Theorem we have

log(J, 1 + ( f P n - = lim

log(1 + JJfP" - 1)dP) J#" - = lim ( log(Jx f P n d P ) lim

Pn n- 00 Pn n-co

n- 00 J,(fP" - 1)dP Pn

= L l o g f d p ,

where { p n } is any sequence of ( 0 , l ) decreasing to 0.

4218

Let {an}n>2 be a sequence of real numbers with lan[ < logn. Consider 00

n=2

a) Prove that this series converges in L1[2, 00) b) Prove that

where the sum on the right is the pointwise limit (you need not prove that this pointwise limit exists).


a) Since

00

The series C ann-2 converges in L1[2, m). n=2

b) Let n

fn(x) = -j+k-5, x 2 2. k = 2

273

cc 00 Then

(i) lim fn(z) = C aklc-" and (ii) Ifn(z)I 5 C lanln-".

By a), 2 H

k=2 k=2 n-w 00

IakIk-" is integrable, and by the Dominated Convergence k = 2

Theorem,

4219

Let S be a bounded Lebesgue measurable set in IR, and let { c n } be a sequence in lR. Show that

where X is Lebesgue measure. (Iowa)

Solution. By the Riemann-Lebesgue Lemma, we have

1 2

= -X(S).

274

4220

(a) Is the function

Lebesgue integrable on the unit square 0 5 x 5 1, 0 5 y 5 l? (b) Compute the repeated integrals in the two orders. (c) Does the integral

exist? (If you use a theorem, be explicit!)

Solution. (a) The function f is not integrable on the unit square, for

If(., Y)ldXdY ) b , Y ? M 2 + Y 2 1 1 )

rdrde

drde

(b) We have

(Iowa)

275

and

4221

Let ( X , M , p ) be a measure space and f E L ( p ) . Evaluate

n-cc lim L n l n ( l+ (!)’) dp.

(Iowa) Solution.

It is easy to show that for any x 2 0

In( 1 + x’) 5 2.

It follows that for any n E lN the function n In( 1 + !$) is dominated by If I and therefore integrable. By the Dominated Convergence Theorem

4222

Suppose that f is a measurable real valued function on lR such that t H

e t z f ( t ) is in L 1 ( E ) for all x E (-1,l). Define the function

276

for all x E (- 1,l). Prove that 'p is differentiable on (- 1,l).

Solution. (Iowa)

Since for any x E (-1, I),

and

we conclude that t H e " t f ( t ) is also in L1(E), E (-1,l). Next we show that

+m

e" f ( t )d t = J_, e t z t f ( t ) d t . dx -m

For any fixed z E (-1, l), for any y > z,

= I(ett - e")tf(t)l, (x < z < y)

= \ e t z " t 2 f ( t ) ( z - x ) ~ (x < z' < z ) t A 2 Iy - z l e ' z X t f(t) , t 2 0 and x < y < kkz

Iy - zlet" t2f( t ) , t < 0 and x < y < A' l zz . 4 By the same method as above, we see that t H et"t2f(t) is integrable. It follows that

and

Evaluate

4223

277

justifying any interchange of limits you use. Hint. First show that (1 + E)n 5 e" for x 2 0.

(S tanfod) Solution.

Since for any n E IV and z 2 0,

i = l

)(l-&)...(l-S)$ Z!

i = l

+

and lim ( 1 + ;)n = e x ,

n-cc

we have x n

( I + --) s e x .

By the Dominated Convergence Theorem

e-"dx = 1. = brn

278

4224

Let ( X , A, p ) be a probability space and f : X + [I, co) a measurable function. Prove or disprove:

(Iowa) Solution.

In f are integrable, and Under the condition that f In f is integrable we conclude that both f and

Indeed. since f ( x ) 5 el ' ' { "f'ln f , otherwise

and 0 5 In f 5 f In f we see that both f and In f are integrable. l n t 5 t l n t for any t > 0,

Since

4225

For i = 1,2, let Xi = nV (the natural numbers), Let Mi = 2N (the g- algebras of all subsets of nV) and let pi be the counting measure. For the function f : X I x X z + B defined by

279

{ ;2--i, j = i, f(i,j) = 2- i , j = is 1,

otherwise.

Compute the iterated integrals

and

How do you reconcile your answers with Fubini's Theorem?

Solution. (Iowa)

For any i E X I , j H f(i,j) is integrable and

f ( i , j ) d p z ( j ) = -2- i + 2-i = 0. J,, Therefore

For any j E X2, i H f ( i , j ) is integrable and

Since f is integrable ( C I f l ( i , j ) = 2), the two iterated integrals exist Li€N

and coincide by the Fubini'i Theorem.

4226

Let (0, p ) be any measure space. For f E L1(Q, p ) , and for X > 0 define

C p ( 4 = P ( { Z E Q I f ( Z ) > XI)

280

and 4(A) = P ( { Z E c2 I f (z ) > -A>).

Show that the functions 'p and II, are Bore1 measurable and that

llflll = J m ( ' p ( A ) + +(A))dA. 0

Hint. As usual, it may be helpful to consider f positive first. (Iowa)

Solution. The measurablity of 'p and + follows from that they are monotone. The function (z, A ) H x~O,~)( l f l (z ) - A) is measurable. On one hand,

but on the other,

By Fubini's Theorem we have

provided that either side exists.

4227

Let f E L2(0 , I). Set ,X

28 1

Show that

for each h, 0 < h < 1 where c is a positive number independent of f.

Solution. (UC, Imine)

We have

( J 1 - h I F ( z + h) - F ( z ) l 2 dz) 0 h

4228

Let f, g E L1(O, 1) and assume that f(z)g(y) = f (y)g(z) for all z, y E [ O , l ] .

282

where A = {(Z ,Y) I 0 52 < Y 511

and dA denotes the planar Lebesgue measure. (Iowa)

283

SECTION 3 SPACE OF INTEGRABLE FUNCTIONS

4301

Let X be a positive measure space of total measure 1. Show that for any [0 , w)-valued measurable function f on X,

is a nondecreasing function of p E (1, m). Under what circumstances is I ( p ) strictly increasing as a function of p?

Solution.

By Holder’s inequality, we have

(Iowa)

For any p , q E (1, w) with p < q , let CY = and = then $ + = 1.

which shows that I ( p ) is a nondecreasing function of p E (1, m). The function I is strictly increasing if and only if f is not almost everywhere equal to the constant function.

4302

Let (X, M , p ) be a fixed measure space, let be positive numbers such that p; > 1, i = 1 , . . . , n, p > 1 and

n 1 1 c,=,. i= l

If, for i E { 1,. . . , n}, fi E CPi ( p ) , must it be the case that

fifi . . * f n E Cp(/l)?

Justify. (Iowa)

284

Solution. Yes, fi * . . fn E LP(p). Moreover

l l f l . * . f n l l p 5 l lf l l lpl . . . IIfnIlp,. (1) P I

In case n = 2, lfilp E L"(p) and If2Ip E L F ( p ) , and by H6lder's inequality,

Assume that the conclusion and the inequality (1) hold for n = k . Then for n = k + l ,

1 PPk 1 < ' 9

- 1 1 1 1 o < -+...+ - = - - - -

Pl P k ?' P k + l pk+l?p

which shows that c+k12p > 1. By inductive assumption f i e . . fk belongs to ppk 1

L Pk+:-P ( P ) and

Then by the case n = 2 we conclude that (fl . . . f k ) f k + l E LP and

which completes the proof.

4303

Let X = { a , b, c}, let M be the cr-algebra of all subsets of X , and let p be the measure determined by

P ( { a ) ) = 0, P ( { b ) ) = 11 and P({.)> = 00.

What are the dimensions of the vector spaces L 1 ( p ) , Lz(p) , and L"O(p)? Jus- tify your answers.

(Iowa)

285

Solution. We have dimL1(p) = dimL2(p) = 1 and dimLW(p) = 2. Since each func-

tion f of L 1 ( p ) or L 2 ( p ) vanishes at c and two functions taking the same value at b coincide in L 1 ( p ) or L 2 ( p ) , we conclude that dimL'(p) = dimL2(p) = 1. Since two functions are equal in L"(p) if and only if they are identical on { b , c } we have dimLW(p) = 2.

4304

lo f d m = 0.

Show that

( Indiana-Purdue) Solution.

Denoting

we have

4305

Let f be a non-negative function in P ( E ) for some 1 5 p < oc) and let T + s = p , T > 0, s > 0. Also let f h ( z ) = f ( z + h).

286

(i) Show that fi f s E L1(R). (ii) Investigate what happens to 1 1 fi fslll as Ihl + co.

(Indiana-Purdue)

= 1, it

Solution.

follows that (i) Obviously f h E Lp(R). From fi E L f , f" E L f , and + P P

f[fS E L1(R).

(ii) We claim that lim 1 1 f i f sII1 = 0.

lhl+m

For any E > 0, take N such that JEc f P < E where E = [ - N , N ] . Now if Ihl > 2 N , then z + h 4 E whenever z E E. Therefore

4306

Let f : R -+ R+ be measurable, and let 0 < T < 00. Show that

(Indiana PPurdue) Solution.

Set p = 1 + f . Then p > 1 and f + & = 1. Since

we have

287

which implies

4307

Let f be a bounded measurable function on (0, l ) . Prove that

(IZZinois) Solution.

For any E > 0,

Then

llfllm =

I

I: Therefore

I l f llm.

288

4308

Let f be a nonnegative measurable function on [0,1] satisfying

Determine those values of p , 1 5 p < 00 for which f E LP and find the minimum value of p for which f may fail to be in P.

(Illinois) Solution.

If 1 5 p < 2, then f E LP. The minimum value of p for which f may fail to be in LP is 2.

Indeed, for any p E [l, 2), m M

it follows that fP is integrable. Let f(z) = & - 1. Then f 2 is not integrable. However for any t > 0

1 2

1 1 < 2. - - ( 1 + t ) 2 l + t

4309

With Lebesgue measure on [0,13, prove that

s = {f E C[O, 11 I l lfl lcc I 1)

is not compact in L1[O, 13.

Solution. (Iowa)

It suffices to show that S is not closed in L1([O, 11). For each n E IN, let

289

4310

Let 'H be Hilbert space L2(0 , 271), with inner product defined by

2= - (u , w) = 1 u(z)w(z)dz, u, w E 'H.

Consider the elements u, E 'H, n = 1 ,2 , . - ., defined by u,(z) = sin(nz) for z E (0,271). Show that

(a) the set {u,}rZl is closed and bounded, but not compact, in the strong (i.e., norm) topology of 2.

(b) u, + 0 as n -+ 00 in the weak topology of X, i.e., for every w E 'H.

lim (un, w) = 0. ,--too


(a) Obviously, the set {u,},",~ is bounded. For any m, n 2 1 with m # n

which shows that {urn I m E IN} is closed. Since it admits no convergent subsequence, it is not compact.

(b) For any w E X, w E L1(0,2r) by Holder's inequality. By Riemann- Lebesgue Lemma,

4311

Let H1 be the Sobolev's space on the unit interval [0,1], i.e., the Hilbert space consisting of functions f E L2[0, 11 such that

290

n=-cc

where

are Fourier coeffents o f f . Show that there exists constant C > 0 such that I l f I l P I Cllflll.


Since for any z E [O,1]

t o o the series C

n=-w

from (1) that

f^(n)e2ninz converges to f(z) both in H1 and in L"O. It follows

I l f l lP I Cllflll.

4312

Let f be a periodic function on lR with period 2a such that f 1 [ 0 , 2 T ~ belongs to ~ ~ ( O , 2 a ) . Suppose

For each h E lR define the function fh by f h ( z ) = f ( z - h). (i) Give the Fourier expansion of fh. (ii) Find the &-norm llfh - f l l z in terms of the a, and h. (iii) Prove that

unless f is constant almost everywhere. (Stanford)

29 1

Solution. (i) We have

n=-w

+w

n=-w

(ii) We have

n = - w

(iii) I f f is constant almost everywhere, then a, = 0, n # 0. It follows that fh = f and 11th - f l l 2 = 0. I f f is not constant almost everywhere, there is an n # 0 such that an # 0. Then

4313

Let f n ( x ) be an orthonormal family of functions in the Hilbert space L2(0, 1). Prove that

for all 2 E [0,1], and that this inequality is sharp (equality) if and only if {f, 1 n = 1,. - - } span a dense subspace of L2(0, 1).

(Iowa) Solution.

By Bessel’s inequality

292

If {fn I R. E nV} span a dense subspace of L2(0 , l), then {fn} is an orthonomal basis of L2(0, 1) and therefore (1) is sharp. Conversely, since

s ~ a n { x ( o , ~ ] I 0 5 z I 1) = s p a n { ~ ( ~ , b l 1 O I a I b 5 I}

is a dense subspace of L2(0, 1) while

is a closed subspace of L2(0, l), containing span{x(o,z] 1 0 5 z 5 l}, it follows

and therefore, {fn} span a dense subspace of L2(0, 1).

4314

I f f is a function on IR, let ft be the translate ft(z) = f ( t + z). Prove that i f f is square integrable with respect to Lebesgue measure, then

lim l l f t - fllL2(nz) = 0. t-+O

(Stanfonl) Solution.

Then ft converges to f uniformly. We have Suppose that f : lR + a' be a continuous function with compact support.

where K = supp(f) and

- [-1,1] = {z - y 12 E K,y E [-1, l]}

is compact. In general, there is a sequence {fn} of continuous functions with compact support converging to f in L2(IR). Then

L

t+O lim l l f t - f l l z 5 S(1lft - fntl l2 + llfnt - fn l l2 + llfn - f l l z )

i t-0 W 2 l l f n - fllz + llfnt - f n l l z ) = 2llfn - fllz.

293

Letting n -, 00, we have lim lift - f l l 2 = 0. t-0

4315

A trigonometric polynomial is a function p on lR of the form

N

n = - N

for some Cn €a', N 2 0. Suppose f is continuous and 21r-periodic on El, and let

Show that for every E > 0, there exists a trigonometric polynomial

N

n = - N

such that Ilf -PI103 < E

and ICnl 5 lan] for all n.

Hint. You may wish to think about harmonic functions on the unit disk. (Stanford)

Solution. Regard f as a continuous function on the unit circle S1 of a'. Then

where a(dz) is the Harr measure on S1 such that a(S') = 27r. Let u be the harmonic extension o f f to the unit disc D = {z 1 IzI 5 1). Then

294

There is an N such that

where r is fixed so that

Put N

n=-N

which is required.

4316

Prove or disprove the following statements: (i) the set of continuous functions on the interval [0,1] is dense in Loo([O, 13). (ii) Lw([O, 11) is a separable metric space.

( St anfo rd) Solution.

(i) False. Since C([O, 11) is separable while Lw([O, 11) is not, C([O, 13) is not dense in Loo([O, 11).

(ii) False. Let

1 n + 1 n < 2 5 -, n E m, f (0) = 0).

1 S = {f E L”([0, 13) I f ( z ) = 0 or 1, -

Then as a subset of L”([O, l]), S is of cardinal N. However, since the distance between any two elements in S is 1, S is not separable.

4317

Let {fn}r=l be a sequence in LP([O,l]) (with Lebesgue measure) and = 1. If lim f n = f in l;P {gn}T=l a sequence in Lq([O,l]), where

and lim gn = g in LQ, is it true that f n g n -+ f g in measure? Justify. +

n-w

n-w (Iowa)

Solution. Yes, fngn -+ fg in measure. Indeed, since

295

we have for any E > 0

where A is the Lebesgue measure.

4318

Let X be a measure space with measure p and suppose that p ( X ) < 00.

Let

S = {(equivalence class) of measurable complex functions on X}.

(Here, as usual, two measurable complex functions are equivalent if they agree a.e.1 For f E S, define

Show that d ( f ,g ) = p(f - g) is a metric on S, and that fn --+ f in this metric if and only if fn -+ f in measure.


Obviously, d(f, g) = d(g, f) 2 0 and d(f, g) = 0 iff & - - 0 a.e., iff f = g in S. For f , g , h E S, we have

296

Therefore d is a metric on S. If fn + f in measure, then -% -+ 0 in measure by the equality

1+f -f

{ 2 E XI I f n - f l > & } = { 2 E XI Ifn - f l 2 l--E ' } ( O < & < l ) . 1 + Ifn - fl -

By the Dominated Convergence Theorem d(fn, f ) -+ 0. If d(fn, f) -+ 0, then

1 + E - - -d(fn,f) -+ 0 as n ---t m.

&

4319

Let g be a measurable function such that s, lfgl < 00 for every f E P(B) (fixed p >_ 1). Prove that there is a constant M such that

all f E Lp(IR).

Solution. ( st unfod)

Define for each n E BV a measurable function gn

gn(2) = { g!x" else. if Ig(z)I 5 n and 121 5 n,

Then gn E Lq(IR) and for any f E L p ( l R ) .

191fl 5 h f l 5 - - * L lsnfl 5 . * . 5 Isfl. Moreover, It follows that the sequence of bounded linear

functionah f H J, f(z)g,(z)dz on P(IR) converges to f H J, f(z)g(z)dz, pointwise. By the Banach-Steinhaus Theorem, f ++ s, f(z)g(z)dz is continuous on P(R) and therefore there is an h E L q ( R ) such that

lim gn f = gf. n-tm

which then implies that g "2. h E Lq(R) .

297

4320

Let S1 denote the unit circle (the set of complex numbers with modulus one, or the real numbers modulo 2a). The convolution of two functions on S1 is

f * g ( a ) = - 2', I'" f(%7(ct - wo. Suppose that f is an element of L2(S1) with the property that its Fourier coefficient

is non-zero for all n E Z. Show that the linear space { f * k 1 k E L2(S1)} is dense in L2(S1).

Solution.

L2(S1). We have

(Iowa)

Let for any n E Z, x,(O) = elne then { ~ ~ } ~ ~ a is an orthonormal basis of

Therefore {f * k I k E L 2 ( S 1 ) } contains {xn I n €27) and therefore is dense in L2( S1).

4321

For functions f , g E L 2 ( R ) , define the convolution f * g by

t o o

f * 9 = J_, f ( y ) g ( z - Y P Y ,

where the integral is with respect to Lebesgue measure. a) Show that f * g E Loo(R) . Do not neglect to check the measurablity of

f *!l.

298

b) Suppose that f has the property that for all g E L2(BZ) the convolution f*g is also an element of L2(IR). Define Tf : L 2 ( R ) -+ L2(BZ) by T f ( g ) = f*g. It is evident that Tf is a linear operator; you need not check this. Show that Tf is a bounded linear operator.

Hint. closed graph theorem. (Iowa)

Solution. Since by Holder's inequality

If(y)g(z - Y)ldY i ([r If(y)l2dY) + (Jt" --oo Id. - YVdY) +

I Ilf l l 2 l l s l l 2 , (1)

f * g(z) is well defined. a) We will show that f * g is uniformly continuous. Indeed,

I r+m I

= 0.

By (l), f * g is bounded and therefore belongs to L m ( B ) . b) Let { g n } be a sequence of L2(IR) such that

gn --t g and Tfgn -+ h in L 2 ( R ) .

Since l(f * 9 n - f * g)(z)l I llfllzllgn - 9 /12 -+ 0, 2 E lR

we see that h = f * g = T f g . By the Closed Graph Theorem, Tf is bounded.

4322

I f f E L 1 ( R ) , define

299

Prove that, for any f E L1(IR), ?(() + 0 as + 00.

Solution. If f is simple, say

( Stanford)

(1)

In general, there is a sequence {fn} of functions of the form (1) such that

Then

Letting n -+ +a, we have

4323

Let f : [0,1] -+ [0, co) be an essentially bounded function, l l f l l o o > 0. Show


For any a with 0 < (Y 0, where X is the Lebesgue measure. For any k E N, by the Dominated Convergence Theorem,

= 0.

Hence

Letting a 1 l l f l l o o , we get that

4324

Let ( X , M , p ) be a measure space for which p ( X ) < 00. Let 1 < p < 00. Suppose that { f k } is a sequence in p ( x ) such that ~upIIfkI(~ < 00 and

k lim f k ( Z ) = f (z ) exists for pa.e . 2. Prove that

n+m

(Indiana) Solution.

If on the contrary lim I l f k - f l l l > 0 there exists asubsequence of {fk}, also

denoted {fk}, such that lim l l f k -fill > 0. Since { l f k l } is bounded in P ( X ) and Lp(x) is reflexive there exists a subsequence { ( f k , I} of {lfkl I k E m}, converging to I f 1 weakly in I ;p(X) .

k + m

k-m

30 1

By the Vitali-Hahn-Saks Theorem, for any E > 0 there exists a 6 > 0 such that for k = 1,2, - . a ,

r r

whenever p ( E ) < 6. For such S > 0 there exists by Egroff’s Theorem a set E such that p ( E ) < S and {fk} converges to f uniformly on X\E. Then

Letting E + 0, we get that

a contradiction.

302

SECTION 4 DIFFERENTIAL

440 1

Let f : [0,13 -+ lR and g : [0,1] x [O , 13 + lR satisfy the inequality

f(Y) - f(.) L g(Y1 Z)(Y - ( for all 2 , Y E (071)). (*)

Assume also that g is nondecreasing in each variable, i.e., u 5 z, v 5 y + g(u, v) 5 g(z, y). Show that lim g(z, y) = +(z) exists except in a countable

set and that for 0 5 z 5 y 5 1 we have Y - + X

Hint. Observe that (*) is equivalent to

(Iowa) Solution.

Let #(z) = g(z, z) then 4 is nondecreasing and

(1) 4(z) L !J(Y,Z) L 4(Y) L g(z1y) L y-’

By (1) we have an inclusion

while the latter set is a t most countable. So lim g(z, y) # 4(z) exists except

in a countable set. Again by (1) f is Lipschitz and therefore absolutely continuous. However, since by (l), f’(z) = 4(z) whenever # is continuous at z,

Y+X

303

4402

Prove that a function f : [0,1] -+ R is Lipschitz, with

for all 2, y E [0,1], if and only if there is a sequence of continuously differentiable functions f n : [O, 11 -+ R such that

(I) [fA(x)l I M for all x E [O, 11; (11) fn(x) + f (x ) for all 1: E [0,1]. Hint. There are several different ways to do this problem; one is to use the

Fundamental Theorem of Calculus.

Solution.

for any x E [ O , 11

(Stanford)

If f is Lipschitz, then f is differentiable almost everywhere and moreover

f(.) = f ( 0 ) + i= f’(x)dz.

Since If‘(z)I 5 M , x E [0,1] there is a sequence { g n } of continuous functions such that

Ign(z)l I M , 2 E [O, 11

and r l

For any n E AT, define

which is required. Conversely, by the Mean Value Theorem


304

4403

Let {fn} be a sequence of absolutely continuous real-valued functions on

(a) f(z) = C fn(z) converges for every 2 E [O, 13.

[0,1] such that 00

n = l

Show that f is absolutely continuous on [0,1]. (nlinois)

Solution. Let

n = l

in L1\O, 11. Then

It follows that f is absolutely continuous.

4404

Assume that f E AC(I) for every I c R. If both f and f’ are in L1(R)

(Indiana-Purdue) show that (i) SR f’ = 0, (ii) f (z) -+ 0 as 121 + 00.

Solution. Since f E A C ( I ) , for every I c R,

305

Since f’ E L1, f E L1, we must have

lim sox f ’ ( t )d t exists, which means lim f(z) exists. Since x-+w x++w

lim f (z ) = 0. Therefore x-+w

f’(z)dz = Iim f(z) - f(O) = -f(o). x++w

In the same way, we have lim f(z) = 0 and x-+-w

Thus 0

f‘(z)dz = J_, f’(z)dz + I’” f ’ ( z )dz = 0. L 4405

Let { f n } C AC([O, l]), fn(0) = 0 for every n. If {f;} is Cauchy ( L l ) , show

( Indiana-Purdue) that there is f E AC([O, 11) such that fn + f uniformly on [0,13.

Solution. Since fn E AC( [0, l]),

f n ( z ) = 1’ fXW.

Thus

So there exists an f E C([O, 11) such that f n -+ f uniformly on [0,1]. L’

Moreover, there exists g E L1 such that f: + g. Then

which implies f E AC( [0, 11).

306

4406

(a) Assume that f E AC(I ) and f’ E L”(I). Show that f is Lipschitz. (b) Show that the following two statements are equivalent. (1) f : I + R is Lipschitz (2) E > 0 3 35 = S ( E ) > 0 such that { I j = [a j , b j ] } c I with


(a) For any zl, 2 2 E I ,

(b) (1) =+ (2) obviously. (2) 3 (1). Take 6 > 0 such that

Suppose that 22 - 2 1 2 6. Take N such that f 5

If(bj)-f(aj)l 5 E for any Ij = [u j , bj] c

< 6. Take { c j } I,

such that x1 = cg < c1 < ... < C N = 2 2 . c, - c,-1 < 6. Then

IIj I 5 6. For any 2 1 , 2 2 , E I , 2 1 < 2 2 , we give the following fact.

Suppose that 22 - 2 1 < 5. Take N 2 Then N l f ( z 1 ) - f(zl)l < 1. So we have

1 I f ( 4 - f(.1)1 I

Therefore we find a Lipschitz constant $.

Let Gn, n = 1 , 2 ,

Let

4407

0, such that 4 < N ( z 2 - 21) < 5.

. be open subsets of [0,1] such that

1 2n GI 3 G 2 3 * * * , lGnl I: -,

307

Show that (9 f AC([O, 11). (ii) If(x’) - f(x”)I 5 MIX’ - x”1, x’, x” E [0,1] iff there exists no such that

(Indiana -Purdue)

(i) For any E > 0, take N such that C & < 5. Set 6 = &. If {(ail bi)}

Gn = 0, n 2 no.

Solution. M

N+1 is a sequence of disjoint open intervals in [0,1] and C(b i - a i ) < 6, then

& & < - & + N C(bi - U i ) < - + - = &,

2 2 2

which means f E AC([O, 11). (ii) Suppose that there exists no such that Gn = 8 for n 2 n o . Then for

2‘‘ > x’, no

If(x”) - f(x’)l = C IG, n [x’,x‘’]~ 5 no1x” - 2’1.

Conversely, suppose that for any natural number K , there is k’ > K , Gkt # 0. Then GK # 0. Take x E GK, 6 > 0 such that (x - S,x + 6) c GK. Take a, b, a < b,

1

K

a , b E (x - 6, x + 6) c GK = Gi. i = l

Then

n= l

K

n = l

K

1

= Klb- al.

308

4408

For what values ‘a’ and ‘b’ is the function

(i) of bounded variation in (-1,l); (ii) defferentiable a t ‘0’. (Iowa)

Solution. The function f is of bounded variation if and only if

To show (l), let us first establish the following equality

provided that either side is finite.

of bounded variation on [ E , 11 and Indeed, since f is continuously differentiable on (0,1], for any E > 0, f is

Then

1

E 1 1

309

Case I. b = 0. Then sup I f (&)I < 00 iff a 2 0, while a 2 0 implies that 0<€<1

sup J' lf'(x)ldx = sup J' l;:!all dx < oc). O < e < l € 0 < € < 1 E

Case 11. b > 0. Then sup I f ( & ) I < 00 iff a + b 2 0, for O<E<'

Since a + 6 = 0 and a + b > 0 imply respectively that

and

the function f' is integrable on (0,1].

that Case 111. b < 0. If a + b 5 0, there is a 6 > 0 such that 0 < x 5 6 implies

d3 I4 Ibl -1bl- - > -. 2 xb - 2

Let N = & + 1. Then 1 b 1

If a + b < 0, since

and

310

I’ If’(.)ld. = +m;

and

we have r l

If a + b > 0, then

and

and therefore f’ is integrable. (ii) f is differentiable a t 0 if and only if

or { b < o { % > l a > 1.

4409

Prove or disprove a) Let f be a real function on [0,27r] satisfying If(.) - f(y)l <: 12 - yI, all

2, y E [0,27r] and f’(z) = 0 a.e. on [0,27r]. Then f must be constant. b) Let f be a real valued function of bounded variation on [ 0 , 2 ~ ] with f’(z) = 0 a.e.. Then f must be constant.


a) The conclusion is ture. Since by the condition that

f is absolutely continuous, f is constant.

31 1

b) is false. Let

f (x) = 0 on [0, T ) and f ( x ) = 1 on [T, 27~1.

Then f is of bounded variation with f’(z) = 0 a.e., but f is not constant.

4410

Let f , g E L1(IR). Show a1

b) if a+h

f(z)dz = c, h-+O

c ) if

then there are constants a , c such that

a+t f ( u + t ) = J, g(z)dz + c a.e.

Can you deduce that f’(x) = g(z) a.e.? State explicitly the theorems you use.

(Stanfod) Solution.

function Assume, without loss of generality, that f and g are real-valued. Since the

f(t)dt = j z f+ ( t ) d t - j z f- ( t ) d t x - L -cc --oo

is of bounded variation, it is differentiable almost everywhere by Lebesgue’s Theorem.

a) For any E > 0 there is a continuous function : lR -+ lR such that

312

We have

Letting c t 0, we see that

and therefore

b) We have by a)

f(' + h, - f ( x ) d Z h

lim h-0 a

= f ( a + t ) - c a.e. t E IR.

c) By a), there is an a E BZ and a c such that

I ra+h f ( z ) d x = c. k; Ja

313

By b) and by the assumption, we have

= 0, a.e. t E IR.

1.e..

It follows that f’(z) = g(z) a.e..

441 1

Let F : (a , b) -+ Dl be measurable. (1) Prove that the following two statements are equivalent: (a) There is an f E L2(a, b) such that

F ( z ) = lz f d m (m the Lebesgue measure);

(b) There is an M > 0 such that

for any finite partition zo < z 1 < .’. < 2, of (a , b). (2) Show that the smallest constant M in (b) is equal to I l f l l ; .

(Iowa)

314

Solution. (1) (a) 3 (b). By the Cauchy-Schwarz inequality

(b) 3 (a) Obviously, for any mutually disjoint intervals (a,, on) of (a , b ) , n E lN, we have

Since

a

F is absolutely continuous, and therefore there is an integrable function f : ( a , b ) ---f E such that

F(x) = LZ f ( t ) d t + c, 2 E (a,b).

We will show that if El, . . . , En are any mutually disjoint measurable sets, then for any E > 0

If Ei's are open sets of (a , b ) , say Ei = , o (aij ,&) (A; at most countable), 3 E A i

315

then by inequality (2)

I 2 It:; f I 2 ;=I j € A i P i j - a i j

5 M .

If Ei’s are compact sets there is for each i a decreasing sequence { E i j } E 1 of

open sets of ( a , b) such that Ei = Eij and Eil n E k l = 0 (i # k ) . Applying

(3) to the mutually disjoint open sets Elj, . . . , Enj and passing to the limits we see that (3) holds for El , . . , En.

In general, there is for each i an increasing sequence { E ; j } E 1 of compact

sets such that u Eij E; and lim p(E;j) = m(E;) . Applying (3) to the

mutually disjoint compact sets Elj, . . . , Enj and passing to the limits we see that (3) holds in general.

00

j = 1

00

j=1 3-00

To show that f is square integrable, it suffices to show that 00

where En = {z In 5 If(z)12 < n+ 1). Let

E,’ = {z I 6 5 f (z ) < -1 and

E, = {x I --< f ( z ) 5 -6}. Then

M

316

5 M .

(2) Define, for each f E L2(a , b ) ,

then 1 1 . 1 1 is a norm on L 2 ( a , b ) by (3). Moreover f E C[a, b], there is

5 l l f l l n by (1). For any E (zip', z i ) (xn = a + k ( b - a ) ) such that

Then

I llf1I2. It follows that l l f l l = Ilfll2 for any f E C[a, b]. For any f E L2(a , b) there is a sequence {fn}rz1 of C[a, b] such that 1 1 fn - f l l 2 ---f 0, so

We have Ilf II = n40O lim l l fnll = I l f n l l 2 = Ilf 112.


31 7

4412

Let X = [0,1] , M the c-algebra of Borel subsets of X. Let a(t) = t2 , p(t) = t3 and define measures ,Y and q5 on M by

p ( E ) = 1 I d a and $ ( E ) = IdP. E

Does 4~ exist? Does 2 exist? Compute the value of the Radon-Nikodym derivatives that exist. Justify.

Solution.

d?

(Iowa)

For any Borel subset E of X , r

p ( E ) = ] 2tdt E

and

We have

+ ( E ) = 1 3t2dt. E

p ( E ) = 0 e 2t = 0 a.e. t E E @ X(E) = 0 e 3 t2 = 0 a.e. t E E e X(E) = 0

e q5(E) = o . It follows that p - 4 - A, where X the Lebesgue measure. We have

and

441 3

Let X be the Lebesgue measure on R2 and /I and v the Borel measures on R2 defined by:

318

Is p << v or u << p? Find the corresponding derivaties if they exist.

Solution. It is true that v << p and it is false that p << v. Because p ( A ) = 0 implies

for all n E N , X((n,2n) n A ) = 0 and therefore p ( ( n , pn) n A ) = 0, i.e., v (A) = 0; however, let A = (g, 21 and then v ( A ) = 0 but p ( A ) = i.

(Iowa)

We show next that

Since

and

we have " 1

n = l du

and dv " 1 - = c $ ~ ( n , i n ] .

n = l dX

Therefore (1) holds.

4414

Let p be the Lebesgue measure on [0, m]. Define

for any Lebesgue measurable subset E of [0, m]. Is p << p2 or/and p2 << pl? If so, find the corresponding derivatives.

(Iowa)

31 9

Solution. Since

r 1

we have

It follows that both p1 << p2 and p 2 << p1. Moreover

From p 2 ( [ 0 , 1 ) ) = 0 and p([O, 1)) = 1, p << p2 does not hold.

4415

Let C#J : lR -+ lR be a bounded, continuously differentiable function with a bounded derivative, and assume g E L(lR). Define

f ( t ) = J dJ(tg(z))dz, t E IR. lR

a) What additional assumption on 4 will insure that f is well defined? (i.e., that q5(tg(.)) E L(lR) for all t E R).

320

b) Under the additional assumption in a) above, show that f is differentiable, i.e., f’(t) exists for all t E IR.

Solution.

an s with 0 < s < 1 such that

(Indiana)

a. Assume 4(0) = 0 then f is well defined. Indeed, for any t , z there exists

Since

b. We will show that

Indeed, for any s, t E R (s # t ) ,

and

The conclusion follows from the Dominated Convergence Theorem.

4416

Let ( f k } denote a sequence of nondecreasing functions defined on (0 , l ) with the property that lim f k ( Z ) = 1 for almost every z E (0 , l ) . Prove that

- lim fi(z) = O for almost every z E (0, I). k + o o

k - r n (Indiana)

Solution. There are sequences {a,} and {b,} such that 0 < a, < b, < 1,

lim (bn - a,) = 1 n-co

321

and lim f k ( a , ) = lim f k ( b n ) = 1

k-+m k - r 00

It follows that lim f; is integrable and k - r m

I’ ks f i ( z ) d z = 0.

Hence lim fL(z) = 0 for almost every z E (0 , l ) . k + c o

322

SECTION 5 MISCELLANEOUS PROBLEMS

4501

Let S c R be a set of real numbers with the property that IS1+. . .+S,I 5 1

( Indiana-Pardae) for every finite subset {,!?I,. . . , Sn} c S. Show that S is countable.

Solution. Since S n (i, n) and S n (-n, -i) must be finite and

S is countable.

4502

Let { I a } , Q E r, be a collection of closed intervals in R. Show that

is countable.

Solution. ( Indiana -Purdue)

Obviously u 1: is an open set. We have aEr

00 u 1: = UI%P,), m a - n = l

where cyn,pnT U I:. For any z € UIa\UI,", there must be (Y such that

x E la, but x can not be in 1:. Take n such that r a m

323

If I , c (an,Pn) , then x 6 Icy, which is a contradiction. Thus we must have

x = on or P,.

Therefore

a a n

which implies u Ia\ u 1: is countable. a a

4503

Show that any infinite set of non-empty, mutually disjoint, open sets in a separable metric space X is countable.

Solution. Let {Ui I i E I } be such an infinite set. For each i E I there is an ai E D

such that ai E Ui, where D is a countable dense subset of X . Then ai # aj

whenever i # j . It follows that the map, i H ai of I to D is injective and therefore I is countable.

(Stanford)

4504

Prove that for almost every x E [0,27r] - lim sin(nz) = 1.

n+co

Solution. Let

( st anford)

2

7F A = {x E (0,27r) I - is irrational}.

Then A is a measurable set of measure 27r. Moreover, for any x E A , - lim sin(nx) = 1.

n-m

Indeed for any x E A, since {Ic: - 21 I Ic, I E a} is a dense subgroup of lR, there are sequences {kn} and { I , } ofZ such that

2 1 lirn (kn- - 21,) = -.

n-m 7~ 2

324

Since f 4 { k : - 2 1 I k,I EZ}, {k,} admits a subsequence {k;} either increasing to t 0 0 or decreasing to -m. If lirn Ic; = $00 then

n-co

lim sin(khz) = lim sin(khz - 2 1 ; ~ ) = 1; n-cc n+cc

Otherwise lirn (-3kA) = +co and n-cc

2 1 lirn ((-3kh)- + 2(31; + 1)) = - n+cc T 2

and therefore

lirn sin(-3kAz) = lirn sin(-3khx + 2(31; + 1 ) w ) = 1. n-+m n-cc

4505

Let f E C ( I ) . Show that there exists a sequence of polynomials { p n } such that pn + f uniformly on I and p l ( z ) 5 p 2 ( z ) 5 . - ' for every z E I .

( Indiana-Purdzle) Solution.

For any n, take a polynomial pn such that

4506

Let f E C([O, 11). Show that there is a sequence of odd polynomials p n ( z )

( Indiana-Purdue)

Suppose that there is a sequence of odd polynomials pn with p , + f . Then

with pn -+ f uniformly on [O,1 ] iff f(0) = 0.

Solution.

f ( O ) = lim pn(O) = 0. n-w

325

Conversely, suppose that f(0) = 0. Set

f (z ) = -f(-z), 2 E [-1,0].

Then f is a continuous function on [-1,1]. Take a sequence of polynomials p, (z ) with p , + f uniformly on [-1,1]. Set

1 2 pn(x) = -[pn(x) - ~ n ( - x ) ] .

Then Pn is an odd polynomial for any n. We have

uniformly on [0, I].

4507

a) Show that the mapping I : C[O, 11 + C[O, 11

1 r+’

is a contracting mapping on C[O, 11, with the supremum norm.

satisfying the conditions: b) Show that there exists one and only one smooth function f on [0,1]

{ &f(z) = ex + zf (x2) , f(0) = 1.

Solution. a) For any f and g in C([O, l]),

(Stanford)

b) By the Banach’s Theorem, there is only one function f E C[O, 11 such that I(f) = f, i.e.,

1 2

f(z) = es + f J, f ( t ) d t . (1)

326

By (1) f is smooth and satisfies the conditions:

f(0) = 1.

If g is another smooth function satisfying (2), then

i.e., I ( g ) = g , by the uniqueness of f, g = f.

4508

Given f : R -+ R bounded and uniformly continuous and {K,} , n =

(i) llKnlll 5 M < 00, n = 1,2,3, . . . . (ii) ~ K , - - + l a s n - + o o . (iii) J12:121,61 lKnl -+ O as n + 00 for all 6 > 0.

Show K , * f -+ f uniformly.

Solution.

there is an N1 such that

1,2,3,. . . , K , E L1 such that

(UC, Irvine)

Take M1 > 0 such that If(z)l < M I for all z E R. For any E > 0, by (ii)

If n > N1. Thus

Take 6 > 0 such that &

lf(z - Y) - f(.)l < 4M holds for all 3: E R any IyI < 6. For the above 6, take Nz such that

327

& & & < - + M . - - z - . 4 4M 2

Set N = max(N1,N2). It follows from (l), (2) that

& & < - + - = & 2 2

holds for all z E R if n > N

4509

Let f : R + (-03, co) be upper semicontinuous and define m : R -+

[-oo,oo) by m(z ) = liminf f(y). Let S = {z I f(z) - m ( z ) 2 1). Y-X

(a) Show S is closed. (b) Show: If I is an open interval contained in S, then m ( z ) = -a on I. (c) Show that S is nowhere dense.

(UC, Irvine) Solution.

(a) For any zo E S", there exists E > 0 such that

f(z0) - m(z0) < 1 - E < 1.

328

Take 6 > 0 such that

for 2 E O(z0,6). Therefore &

m(z) = liminf f (y) 2 rn(zo) - Y+Z

holds for z E O(z0,S). Thus & & f(.) - m(2) < f(20) + - - m(.o) + - < 1 - & + & = 1 2 2

if z E O(zo,6), i.e., 2 E S". So S" is open. (b) Suppose that there is zo E I such that rn(z0) > -m. As in (a), we can

find 6 > 0 such that m(z) > rn(z0) - i for 2 E O(z0, 6) c I . By the definition of rn(z), there must be F E O(z0,6) such that f(5) < rn(z0) + i. Therefore f(:) - m(5) < 1, i.e., 5 6 S which contradicts % E I c S.

(c) Suppose that S is not nowhere dense. Since S is closed, there is an open interval I c S. From (b), m(z) = -m for z E I . Denote

Al = {Z E I,f(.) < -1).

A1 is a non-empty open set. Take an open interval I1 such that 71 c A l . In the same way, we can take an open interval I2 satisfying 7 2 c I1 and f(z) < -2 for z E 72. By induction, we obtain a sequence of open intervals { I n } such that 7, c In- l , f(z) < -n for z E 7,. Obviously f (z ) = --oo for 2 E n?,, which is a contradiction.

4510

Prove that any topological metric space is homeomorphic to a bounded metric space.


Suppose that ( X , d ) is a metric space. Define another metric d o n X by

It is easy to show that the identity mapping of X is a homeomorphism of ( X , d) to ( X , Z ) .

329

4511

Let M be a metric space with distance function d, suppose A : M +. M is a distance nonincreasing periodic map of order 3, i.e.,

A o A o A = Id and d(Az , Ay) 5 d(z, y).

(i) Show that A is a continuous bijective isometry. (ii) Give an example of a complete metric space M and an isometry A on

M , periodic of order 3, which has exactly two fixed points.

Solution.

isometry and therefore continuous.

(Stanford)

(i) Since A-' = A2, A-' is distance nonincreasing, too. Therefore A is an

(ii) Let M be the 2-dimensional sphere

M = { ( z , z) E C x 2R I 1 . ~ 1 ~ + z2 = 13.

The isometry A : M -+ M defined by

A(z , z) = (e;* 'z, z), ( z , z) E M ,

is a periodic map of order 3, having exactly two fixed points ( 0 , l ) and (0, -1).

4512

Let H be a Hilbert space and let f : H -+ B be a continuous convex function such that f (zn) -+ oa whenever llznll ---t 00. Prove that f attains a minimum.

Solution. (SVNY, Stony Brook)

Let (Y = inf f ( H ) . There is a sequence {zn} of H such that

lim f(zn) = a . n-co

If SUP llznll = $00, n

there is a subsequence {Z~,}F=~ such that l/znk ( 1 ---f +m, then

a = lim f (zn,) = +00, k-rco

330

a contradiction. So sup llzn[l < +m. By the weak compactness of the closed

ball of H , there is a subsequence { z n k } converging weakly to a point z E H .

Since

n *

For any p > a there is an N E lhT such that for any k > N , f (xnk) < 0.

X E W

cov{z,, 1 k > N } = 11.11 coV{znk I k > N }

and for any y E cov{z,, I k > N } , f (y ) < ,B, one has f (z ) 5 p. Therefore

It follows that a is finite and f attains its mininum at z.

4513

A is the subset of L1(BZ) consisting of all functions f satistying If(z)I 5 1 a.e. on aZ. Prove that A is closed in the norm topology of L1(lR).

Solution. ( StunfoTd)

If {fn} is a sequence of A such that f = lim f,, exists in L1(BZ). Then {fn} converges to f in measure and therefore by F.Riesz Theorem there is a subsequence {fnk} converging to f almost everywhere. I t follows that If(z)I 5 1 a.e. on lR. So A is closed.

n-+m

4514

Let A be a bounded linear operator on Hilbert space H . Recall that the adjoint A* is the unique bounded linear operater on H such that (Az ,y ) = (z, A*y) for all z,y E H .

(a) Show that IIA*ll = IIAll, where IlAll is the norm of A. (b) Show that AA* - A* A cannot be the identity on H. (You may wish to

use (a) prove this.)


(a) Indeed, since

llA*11 = SUP llA*~ll = SUP SUP I(x7A*y)I IlYll51 I l Y l l 5 1 114151

l lY l l 51 114151 - - SUP SUP I(Az,y)I 5 IlAll

33 1

and a fortiori IlAll = 11A*\\. (b) First, we will show that

Indeed equalities (1) follow from the following inequalities

If for some bounded linear operator A on H , AA* - A*A = I then

I t follows that I)A))2 = JJAJ12 + 1, a contradiction.

4515

Suppose that A, B c IR are Lebesgue measurable, with m(A) > 0, m ( B ) > 0. Show that

contains an interval of positive Lebesgue measure.

Solution. (Indiana)

Assume without less of generality that A and B are compact sets. Since

= m(A)m(B) > 0,

332

there exists an zo E IRsuch that m((zo-B)nA) > 0. Since z H m((z-B)nA) is continuous, the set {z I m((z - B ) n A > 0) is nonempty and open. Then the conclusion follows from the following inclusion

{Z I m((z - B ) n A ) > 0) C A + B.

4516

Let X be a compact Hausdorff topological space and let p be a finite regular Bore1 measure on X . Is it true that if f : X -+ IR is p-measurable then there exists a sequence {fn}~'l of continuous real-valued functions such that lim f, = f a.e. [p]? Justify.

Solution. Yes.

n-0 (Iowa)

For any n E lN there is a compact subset F, of X such that

p(X\F,) < and f is continuous on F,. Let X , = u Fi. Then (1) i = l

00

p(X\X,) < i, ( 2 ) f is continuous on X,, and (3) p

is for each n E lN a real-valued continuous function f, : X -+ lR such that f, = f on X,. Then lim f, = f a.e. b].

n-co

Part V

Complex Analysis

335

SECTION 1 ANALYTIC AND HARMONIC FUNCTIONS

5101

True-False. If the assertion is true, quote a relevant theorem or reason; if false, give a counterexample or other justification.

(a) if f(z) = u + iv is continuous at z = 0, and the partials u,, uy, v,, vy exist at z = 0 with u, = vy and 1-ly = -21, at z = 0, then f'(0) exists.

(b) if f(z) is analytic in R and has infinitely many zeros in R , then f E 0. (c) if f and g are analytic in R and f(z) . g(z) E 0 in R, then either f E 0

(d) if f(z) is analytic in R = { z ; Rez > 0 } , continuous on with If(iy)I 5 1

(e) if C a n z n has radius of convergence exactly R, then Cn3a,zn has

(f) sin f i is an entire function.

or g 0.

(-co < y < +co), then If(z)I 2 1 ( z E R).

radius of convergence exactly R.

(Indiana-Purdue) Solution.

(a) False. f satisfies Cauchy- Riemann equations at z = 0, but f'(0) doesn't exist.

(b) False. A counterexample is f (z) = sin&. f is analytic in R = { z : IzI < l}, and has zeros z = 1 - 6, n = 1,2, . -- . But f is not identically zero in R.

(c) True. If neither of f and g is identically zero in R , then both f and g have at most countably many zeros in R, and the zeros have no limit point in R. Then f(z) . g ( z ) is not identically zero in R.

(d) False. A counterexample is f(z) = e", which is analytic in R, and continuous on a with If(iy)I E 1. But f (z ) is not bounded in R.

(e) True. Because lim G= 1, it follows from

A counterexample is f ( z , y ) = m.

n-+m

336

(f) False. sin f i is not analytic a t z = 0. Actually, z = 0 is a branch point of sin &.

5102

(a) Let f(z) be a complex-valued function of a complex variable. If both f ( z ) and z f ( z ) are harmonic in a domain Q, prove that f is analytic there.

(b) Suppose that f is analytic with If(z)I < 1 in IzI < 1 and that f ( i -u) = 0 where u is a complex number with 0 < la1 < 1. Show that If(0)l 5 u2. What can you conclude if this holds with equality.

(c) Determine all entire function f that If’(.)[ < l f ( z ) I . (Stunf07-d)

Solution. (a) It is well known that the Laplacian can be written as

and

that

which implies that f ( z ) is analytic in R.

then F ( z ) is analytic in (1.1 < 1). When IzI = 1,

1-7iz 1+az

hence

337

which implies that IF(.)[ 5 1 for IzI < 1. Take z = 0, we obtain

I f (0) l I bI2. When it holds with equality, we have F ( z ) elel which is equivalent to

z - u z + a f (z ) = ,a’-. -. 1-Tiz 1 + z z

(c) From lf’(z)l < lf(z)I, we know that f has no zero in a‘, which implies that fIIl is also an entire function. It follows from I $$I < 1 that % = c, IcI < 1. Integrating on both sides, we obtain logf(z) = cz + d. Hence f(z) = c’e‘”, where c and c’ are constants and JcI < 1.

f(2)

5103

Let G be a region in a‘ and suppose u : G -t lR is a harmonic function. (a) Show that 2 - i% is an analytic function on G. (b) Show that u has a harmonic conjugate on G if and only if - ie has

a primitive (anti-derivative) on G.

Solution. (a) Let

(Indiana)

Because u is a harmonic function, we have

We also have

ap aQ a Z u

ax a y a x 2 a y 2

a p aQ a Z U a z u a y ax axay azay

- -+ ~ = 0.

= 0. -+-=-- -

So P ( x , y) and Q ( x , y) satisfy the Cauchy-Riemann equations, hence

au au ax a y

P + i Q = - - i -

is analytic on G. (b) If 2 - ig has a primtive, then for any closed curve c c G, the integral

338

It follows that dU du

- K d x + z d y = 0

holds for any closed curve c c G. Hence we can define a single-valued function

where zo, z E G, and the integral is taken along any curve connecting zo and z in G. Because av au av - au

a x dy ’ a y a x ’ _ _ - - - _ - -

we know that v(x, y) is a harmonic conjugate of u(x, y) on G. On the contrary, if u has a harmonic conjugate v on G, then

av a V d U dU dv = -dx + -dy = --dx + -dy. ax a Y ay ax For any closed curve c C G , we have

= l d ( u + iv) = 0.

Hence % - i k has a primitive sz: aY

5104

Suppose that u and v are real valued harmonic functions on a domain R such that u and v satisfy the Cauchy-Riemann equations on a subset S of R which has a limit point in R. Prove that u + iv must be analytic on 0.


- iau aY aY are analytic functions on a. The reason lies on the fact that the real and imaginary parts of fi and f2 satisfy the Cauchy-Riemann equations respectively (see 5103 (a)).

Because u and v are harmonic functions, fi = 2 - ik and f2 =

339

By the assumption of the problem,

au av du av - - - _ _ - _ - - ax d y ’ d y ax

when z E S c Q. Hence

fi = & bu - i& au = if 2 = i ( g - i $ )

when z E S. Because the subset S has a limit point in Q, by the uniqueness theorem of analytic functions, we know that f1 = if2 holds for all z E Q. It follows from fi = i f 2 for z E R that

for z E Q, which implies that u + iw is analytic on Q.

5105

Let Q = [0,1] x [0,1] c a‘ be the unit square, and let f be holomorphic in a neighborhood of Q . Suppose that

f (z + 1) - f (z ) is real and 2 0 f (z + i) - f (z ) is real and 2 0

for z E [O,i] for z E [0,1].

Show that f is constant.

Solution.

theorem, we have

(Indiana)

Because f is holomorphic on the closed unit square Q, by Cauchy integral

As

340

for 0 5 z 5 1 and f(1 + Yi) - f ( Y 4 2 0

for 0 I y 5 1, by comparing the real and imaginary parts in the above identity, we obtain that f(z+i) = f (z) for 0 5 z 5 1 and f ( l + y i ) = f (yi) for 0 5 y _< 1. Hence f ( z ) can be analytically extended to a double-periodic function by

f (z ) = f ( z + 1) = f ( z + i ) ,

which is holomorphic in a‘ and satisfies

lf(z)I I %${If (.)I1 <

This shows that f (z ) must be a constant.

5106

Let f be continuous on the closure 3 of the unit square

s = { z = z + i y E @ ’ : 0 < 2 < l , o < y < l},

and let f be analytic on S. If Rf = 0 on sn ({y = O } U {y = l)), and if If = 0 on Sn ({z = 0) U (2 = l}), prove that f = 0 everywhere on S.

Solution. Define F ( z ) = si f(z)dz, where the integral is taken along any curve in

which has endpoints 0 and z. Then F ( z ) is analytic in S and continuous on S. For z E as, we choose the integral path on dS and consider the differential form f(z)dz in the integral. Let f = u + i w , then

(Indiana)

-

f(z)dz = (udz - vdy) + i(wdz + udy) .

On 3 n ({y = 0) U {y = 1)) we have u = 0 and dy = 0, and on sn ({z = 0) u {z = 1)) we have v = 0 and dz = 0. Hence we obtain Re(f(z)dz) = 0 on dS which implies ReF(z) = 0 when z E dS.

Let G ( z ) = e F ( * ) . Then G(z) is analytic in S and IG(z)I = 1 when z E 85’. Because G(z) has no zeros in S, so l/G(z) is also analytic in S and 11/G(z)l = 1 when z E dS. Apply the maximum modulus principle to both G ( z ) and l /G(z), we obtain IG(z)I 1 for z E S, which implies that G ( z ) is a constant of modulus 1. It follows from G(z) = eF(’) that F ( z ) is also a constant. Hence f (z ) = F’(z) G 0.

341

5107

(a) Find the constant c such that the function

C -- 1 f ( z ) = z 4 + z 3 + 2 2 + 2 - 4 z - 1

is holomorphic in a neighborhood of z = 1.

closed disk { z : IzI 5 1).

Solution.

(b) Show that the function f is holomorphic on an open set containing the

(Iowa)

(a) As

1 24 + 23 + 22 + z - 4

lim(z - 1) - 2 4 1

1 = lim

2 4 1 (z4 + 23 + 22 + z - 4)’ 1 = lim

2-1 4z3 + 3z2 + 2z + 1 1

10’ - - -

1 we know that z = 1 is a simple pole of a4+as+2z+2-4 with residue equal to &. Hence when c = A, f (z ) is holomorphic in a neighborhood of z = 1.

(b) When IzI 5 1, we have

3 l z4 + z 3 + z 2 + - 41 2 4 - lz4 + 2 + z 2 + z~ 2 4 - 1 4 4 - l z 1 - lz12 - l z 1 1 0,

and the equalities hold if and only if z = 1, which shows that z = 1 is the only zero of z4 + z3 + z2 + z - 4 in { z : IzJ 5 l}. By (a), we obtain that f ( z ) has no singular point in { z : IzI 5 l}, hence f (z ) is holomorphic on an open set containing { z : IzI 5 I}.

5108

Let P ( z ) be a polynomial of degree d with simple roots z1, z2, . . . , zd. A “partial fractions” expression of $ has the form:

342

(a) Give a direct formula for cn in terms of P. (b) Show that 2 really has a representation of the form (*). (c) Give a formula similar to (*) that works when z1 = z2 but all other

p(.)

roots are simple. (Courant Inst.)

Solution.

(4 1 - z - zn

c n = lim z-+.zm P ( z ) P’(zn)’

which is the residue of -2- a t z = z,. P ( . )

(b) Let

Then f ( z ) is analytic o n c and lim f ( z ) = 0. By Liouville’s theorem, f(z) is identically equal to zero, hence

I 4 0 0

(c) Denote the Taylor expansion of P ( z ) at z = z1(= z2) by

00

P(2) = c an(z - Z1)n.

Then the Laurent expansion of 2 a t z = z1 is P ( Z )

Hence 1 has the form: P ( , )

343

5109

Suppose f is meromorphic in a neighborhood of D (D = { ( z ( < 1)) whose only pole is a simple one at z = a E D. If f ( a D ) C IR, show that there is a complex constant A and a real constant B such that

Az2 + Bz + z f(.) = ‘(z - a ) ( l - Tiz) *

(Indiana) Solution.

Assume that the residue o f f at z = a is Al. Define

Ai Aiz z - a 1 - z z

g(z) = f (z ) - - - -.

It is obvious that g ( z ) is analytic on D and g(aD) C lR. By the reflection principle, g ( z ) can be extended to an analytic function on the Riemann sphere z, hence g ( z ) must be a constant. Suppose g ( z ) B1, then B1 is real and

Az2 + Bz + z ( z - a ) ( l - Tia)’

- -

where A = 2 1 - ZB1, B = -(ZAl + a z l ) + B1(1+ (at2) E lR.

5110

Let K1, K2, . . . , Kn be pairwise disjoint disks in C, and let f be an analytic

function inC\ U Kj. Show that there exist functions f1, fz,... , fn such that n

j=1 (a) f j is analytic in C\Kj, and

(b) f (z ) = C fj(z) for z €C\ U Kj. n n

j=1 j=1 (Indiana)

Solution. Assume K1 = { z ; (a - z l ( 5 rl}. Choose c1 > 0 sufficiently small, such that

n

C1 = { Z; TI < ) Z - ~1 < + ~ 1 ) C C\ U Kj * j = 1

344

In C1, f (z ) has the Laurent expansion

+m

f(z) = c u p ( z -.I)?

k = - m

Set 0

fl(Z) = c u p ( z - z1)5 k = - o o

f l ( z ) is analytic in @\K1. Because f (z ) - fi(z) has an analytic continuation n

to K1, f (z ) - fl(z) is analytic in @\ u Kj. j = 2

Assume K2 = { z ; Iz - 221 5 7-2). Choose ~2 > 0 sufficiently small, such

Laurent expansion

(2) uk ( z - ~ 2 ) ~ . f2(z) is analytic in@\KZ. Because f(z) -

fl(z)-f2(z) has an analytic continuation to K2, f(z)-fl(z)-f2(z) is analytic

in @\ ij ~ j .

0

k = - m Set f2(z) =

j = 3

Repeat the above procedure n - 1 times, we get a function f (z ) - fl(z) - fZ(Z) - * * * - f n - l ( z ) , which is analytic in @\Kn. Set

Then we have n

j=1

where f j (z) is analytic in@\Kj, and the above identity holds for z E @\ U Kj. n

j = 1

5111

Recall that a divisor Df of a rational function f(z) on @ is a set { p E @ u {m}}, consisting of zeros and poles p of f(z) (including the point co),

345

counted with their multiplicities np E 2. Let f and g be two rational functions with disjoint divisors. Prove that

(SUNY, Stony Brook)

Let pi (i = 1,2, . - 1 , n) be all the zeros and poles of f(z) with multiplicities npi respectively. I t should be noted that p; is a zero of f when npi > 0 and a pole of f when npi < 0. By the property of rational functions, we have

C npi = 0. Similarly, let q j ( j = 1 ,2 , . . . , m) be all the zeros and poles of g ( z )

with multiplicities mqj respectively, then we have

Solution.

n

i = l m

j=1 mqj = 0.

First we assume that the point 00 is not a zero or a pole o f f or g, then f and g can be represented by

n

f (z ) = A JJ(z - p i ) n p i

and m

Then

n n n m

and

m m m n

346

In case the point 00 is a zero or a pole of f or g, we may assume p , = 00

without loss of generality. Then

n-1

f ( z ) = A n ( z - i = l

and m

g ( z ) = B n ( ~ - 4j)"'j.

j=1

m

j=1 Since C mpj = 0, we may assume that g(p,) = g(m) = B. Hence

n- 1

i = l

n-1 m

and

i = l j=1

which completes the proof of the problem.

5112

Let f ( z ) be the "branch" of logz defined off the negative real axis so that

(a) Find the Taylor polynomial of f of degree 2 at -4 + 3i, simplifying the

(b) Find the radius of convergence R of the Taylor series T ( z ) of f ( z ) at

f( 1) = 0.

coefficients.

-4 + 3i.

347

(c) Identify on a picture any points z where T ( z ) converges but T ( z ) # f(z), and describe the relationship between f and T at such points. If there are no such points, is this something special to this example, or a general impossibility? Explain and/or give examples.

(Minnesota) Solution.

(a) When z is in the neighborhood of zo = -4 + 3i, we have

f (z ) = logz = l0g[(-4 + 3i) + (Z + 4 - 3i)]

I [ -4 + 3i z + 4 - 3 i

log( -4 + 3i) + log 1 + 3 z + 4 - 3 i 5 - 4 + 3 i

log 5 + i(7r - arcsin -) + 2 1 z + 4 - 3 i

-5 ( -4+3 i ) +...*

Hence the Taylor polynomial of f of degree 2 at -4 + 3i is

co + c1(z + 4 - 3i) + cg(z + 4 - 3 4 2 ,

where

3 co = log 5 + i(7r - arcsin -),

4 + 3i 25 '

5 -- c1 =

and 25 + 24i

c2 = - 1250 '

(b) Denote the Taylor series of f(z) at -4 + 3i by T ( z ) . Because log z has only z = 0 and z = 00 as its branch points, and has no other singular point, the radius of convergence R of T ( z ) is equal to the distance between z = -4 + 3i and z = 0. Hence R = 5.

(c) Denote the shaded domain shown in Fig.5.1 by a. When z E Cl = { z : Iz + 4 - 3i( < 5,Imz < 0 } , T ( z ) # f(z). It is because T ( z ) in R is the continuation of logz at -4 + 3i in the disk { z : Iz + 4 - 3il < 5}, while f (z) in R is the continuation of log z at -4 + 3i in the slit plane @\(-m, 01. Hence

348

the difference is 2ai, i.e., T(z) = f (z ) + 27ri.

Fig.5.1

5113

Let f be the analytic function defined in the disk A = { z : Iz - 41 < 4) so that f ( z ) = z i ( z + 1)i in A and f(z) is positive for 0 < 2 < 8. An analytic function g in A is obtained from f by analytic continuation along the path starting and ending at z = 4 (see Fig.5.2). Express g in terms of f .

(Indiana)

Fig.5.2

Solution.

z goes along r from the start point to the end point by A,$(.). Then Denote the closed path in Fig.5.2 by I?, and denote the change of $ ( z ) when

g ( z ) = Ig(z)leia%dz) = ~f(.) I,l(argf(z)+Arargf(zr)).

We have 1 1 1 1 -Arargz + -Ararg(z + 1) = -(27~) + - ( - 2 ~ ) 3 2 3 2 Arargf(z) =

Hence

349

5114

Define eJ; - e - f i

sin& * f ( z ) =

(a) Where is f single-valued and analytic? (b) Classify the singularities o f f . (c) Evaluate 4x,=25 f(z)dz.

(Indiana) Solution.

(a) It is known that z = 0 and z = 00 are the branch points of function &. Let r = { z : IzI = T } , and when z goes along I’ once in the counterclockwise sense, f i is changed to -fi, while f(z) is changed to

e - 6 - e& e f i - e-J; - -

sin(-&) sin f i which is still f(z). Hence z = 0 and z = 00 are no longer the branch points of

When z is in the small neighborhood of z = 0, f(z) can be represented by f ( z ) .

03

2 c n = O -zn -

00 - > c O n z n

(%+I)! n = O

which implies that z = 0 is a removable singular point of f (z) . It is obvious that z = n2R2 (n = 1 , 2 , . . .) are poles of f (z) . Hence f ( z ) is single-valued and analytic inC\{z = n2r2 : n = 1 , 2 , . . . } .

(b) We have

which shows that z = n2r2 are simple poles of f (z ) with residues

2 n ~ ( - l ) ~ ( e ~ = - e-nn).

350

As to z = 00, it is the limit point of the poles of f(z), and hence is a

(c) f (z ) has only one pole z = 7r2 in the disk { z : IzI < 25). Hence non-isolated singular point of f(z).

f(z)dz = 2 ~ i R e s ( f , r2 ) L < 2 5

5115

Let fl be the plane with the segment (-1 5 z 5 1 , y = 0) deleted. For which of the multi-valued functions

(a) f(z) = (b) g ( z ) = 1

can we choose single-valued branches which are holomorphic in fl. Which of these branches are (is) the derivative of a single-valued holomorphic function in 0. Why?


Let I? be an arbitrary simple closed curve in a, and denote by A,r#J(z) the change of r#J(z) when z goes continuously along I' counterclockwise once. It is known that f and g can be represented by

and

- 1 - ,{-+log(l+z)-+log(l--l)} g ( z ) - 4- -

1 1 2 2

Because Ar[argz - -arg(l + z ) - -arg(l - z ) ] = 0

and

0 -27r

(-1 5 z 5 1 , y = 0) not inside r (-1 5 z 5 1 , y = 0) inside I?

1 1 2 2

Ar[--arg(l + z ) - -arg(l - z ) ] =

351

we have A r f ( z ) = 0 and Apg(z) = 0. Hence both f (z ) and g ( z ) have single- valued branches which are holomorphic in Q, and each of f and g has two single-valued branches.

In order t o know which of f and g has a single-valued primitive in R, we consider the integrals 1, f(z)dz and 1, g(z)dz. If the segment (-1 5 z 5 1, y = 0) is not inside I?, it is obvious that s, f (z)dz = 0 and Sr g(z)dz = 0. If the segment (-1 5 z 5 1,y = 0) is inside I?, we consider the Laurent expansion of f and g about z = 03:

1 1 a2 a4

2 2 f (z ) = *i(l - -)-5 = f i (1 + 7 + 7 + * * .) ,

1 ( z 2 3 z 1 b3 b5 - + - + Ti +... g(2) = *-(1- -)-T =fi i 1 1

z 2 2

It follows that sr f(z)dz = 0 and lrg(z)dz = f 2 a . Hence we obtain that both of the single-valued branches of f are the derivatives of single-valued holomorphic functions in Q, and the primitives are szt f(z)dz + c, where the integral is taken along any curve connecting zo and z in 0. But neither of the branches of g is the derivative of a single-valued holomorphic function in a.

5116

(a) Let D c C be the complement of the simply connected closed set (eetie 1 6 E R} U (0). Let log be a branch of the logarithm on D such that loge = 1. Find log el5. Justify your answer.

(b) Let y denote the unit circle, oriented counterclockwise. By lifting the integration to a n appropriate covering space, give a precise meaning to the integral J'(1og ~ ) ~ d z and find all possible values which can be assigned to it.

Solution. (a) The set {ee+ie I 6 E IR} U (0) is a spiral which intersects the positive

real axis a t {e2nn : n = 0,*1,*2,. . a ) . The single-valued branch of logz is defined by loge = 1. Hence loge15 = loge + A r l o g z , where I' is a continuous curve connecting z = e and z = el5 in D and A r log z is the change of log z when z goes continuously along r from z = e to z = el5. I t follows that Ar logz = Aplog IzI + iArargz, and Ar log IzI = 15 - 1 = 14. Because e E (e', e2=), el5 E (e4A, e6=), we know that when I' connects e and el5 in D, Arargz must be 4a. Hence log el5 = 1 + (14 + 4ai) = 15 + 4ai.

(b) Define the lift mapping by w = logz which lifts the unit circle y one-to- one onto a segment with length 2ir on the imaginary axis of w-plane. Because

(HUT7NZrd)

352

both the starting point of y and the single-valued branch of log z on y can be arbitrarily chosen, the segment on w-plane can be denoted by [it,i(t + 27r)), where t can be any real number. Hence we have

i ( t + 2 * ) i( t+27r)

weWdw - 21t i ( t + 2 x )

J ( W 2 d z = lt w2ewdw = (w2ew)lit 7

i ( t+la)

+ 2 1 t ewdw i ( t + Z U ) = eit(-47rt - 47r2) - (2wew)lit

-47~(t + ?r + i)eit = 47r(t + 7~ + i )e i ( t+7r) , =

which implies that the set of values being assigned to the integral J (log ~ ) ~ d z is a spiral (4-43 + i)ezs : s E R}.

7

5117

Find the most general harmonic function of the form f(lzI), z E C\O. Which of these f(I.1) have a single valued harmonic conjugate?

Solution. Because f ( I . 1 ) is harmonic, we have reason to assume that the function f

(with real variable t ) has continuous derivatives f’(t) and f”(t). Note that the Laplacian

(Indiana)

a2 a2 a2 a x 2 a y 2 azaz7

A = - - - + - - - = 4 -

and

we obtian

where t = IzI. This differential equation is easy to solve, and the solution is f ( t ) = Q log t + 0, where (Y,P are two real constants. Hence the most general harmonic function of the form f( 1.1) in C\O is Q log IzI + 0.

Since log IzI has no single-valued harmonic conjugate inC\O, we know that when f( 1.1) has a single-valued harmonic conjugate in C\O, it must be a constant.

353

5118

Consider the regular pentagram centered at the origin in the complex plane. Let u be the harmonic function in the interior of the pentagram which has boundary values 1 on the two segments shown and 0 on the rest of the boundary. What is the value of u at the origin? Justify your claim.

Solution. Denote the interior domain of the pentagram shown in Fig.5.3 by D, and the

ten segments of the boundary by I l l I z , . . . , 110, put in order of counterclockwise.

(Stanford)

Fig.5.3

Then denote the harmonic function on D with boundary values 1 on l k and 0 on the rest of the boundary by u k ( z ) , k = 1 ,2 , .. . , 10. By the symmetry of domain D, we have

It follows from i n

354

5119

Suppose G is a region in C, [0,1] c G, and h : G + lR is continuous. h l ,qp l ] is harmonic, does this implies that h is harmonic on G?

Solution. (Iowa)

The answer is No. A counterexample is h ( z ) = Red-, where the single-valued branch

of d m is chosen by d m l z = z = 4. Since d m is analytic in C\[O, 11, h ( z ) is harmonic there. When 0 5 2 5 1,

lim & z - i j = & c T ) i ,

lim d m = -4-i.

z=x+yi+z

Y>O

z=x+yi-+x

Y<O

Hence h ( z ) = 0 when z = 2, 0 5 2 5 1, and h ( z ) is continuous onC. But h(z) is not harmonic on&, because z = 0 and z = 1 are branch points of d m .

Remark. If the problem is changed to h : G -+ C is continuous and hl~\[o,l] is holomorphic, then h must be holomorphic on G.

5120

Let 7 be an arc of the unit circle. Suppose that u and v are harmonic in D = { z : IzI < 1) and continuously differentiable on D u y. If the boundary values satisfy u = v on y and the radial derivatives satisfy = on y, prove that u = v in D.

Solution. Let u* be a conjugate harmonic function of u in D and v* be a conjugate

harmonic function of v in D. We know that a variation of Cauchy-Riemann equations for f = u + iu* and g = v + iv* are

(Indiana)

dU* - d u du* au r-=- - dr ae ae --'dr

and

355

It follows from the continuous differentiability of u and v on D U y that u* and v* can be continuously extended to D U y and then are also continuously differentiable on D U y. Let zo be a fixed point on y, and for z E y denote the subarc of y from zo to z by yz. Without loss of generality, we may assume that u * ( z o ) = v*(zo) = 0. Then for z E y,

= lz :dB = v*(z).

Hence we obtain two functions f = u + iu* and g = v + iv* which are analytic in D and continuous on D Uy, such that f = g on y. Let F = f - g. Then by the reflection principle, F can be analytically extended to an analytic function on D u y U D*, where D* = {z : IzI > 1). Since F = 0 on y, we obtain F 0 on D U y U D*, which implies u = v in D.

5121

Use conformal mapping to find a harmonic function U ( z ) defined on the unit disc { Iz1 < 1) such that

+1 f o r O < d < a -1 for 7r < 8 < 27r.

lim U(reie) = r + l -

Give the correct determination of any multiple-valued functions appearing in your answer.

(Courant Insf.) Solution.

It is easy to know that w = -is is a conformal mapping of the unit disc D = { z : IzI < 1) onto the upper half plane H = {w : Imw > 0). The boundary correspondence is that the negative real axis {w : --oo < w < 0) corresponds to the arc I’l = { z = eie : 0 < 8 < T) and the positive real axis {w : 0 < w < +a} corresponds to the arc rz = { z = eie : T < d < 2 ~ } .

It is well known that u(w) = targw - 1 is a harmonic function in H and assume $1 on the negative real axis and -1 on the positive real axis. Hence

z S 1 2 z + l z - 1 T z - 1

U ( z ) = u(-i-) = -arg(-) - 2,

where the single-valued branch of arg( 2) is defined by arg( %)lr=o = a, is a harmonic function in D = { z : It1 < 1) with the boundary values $1 on and -1 on I'2.

as follows: Remark. This problem can be solved directly from the Poisson formula

1 = -A,,{2arg(C - z ) - argc) - 1

2 z + l - -arg- -2 .

R 2 - 1

A

-

5122

Determine all continuous functions on { z E C : 0 < 1z1 5 l} which are harmonic on { z : 0 < IzI < 1) and which are identically 0 on { z EC : IzI = 1).


Suppose u ( z ) is a continuous function on (0 < IzI 5 1) which is harmonic on (0 < IzI < 1) and identically zero on { Iz I = 1). Let * d u = -u,dz + u,dy and A = hzl=r*du, where A is a real number not necessarily zero. Denote w(z) = Jz * d u , then w(z) is the conjugate harmonic function of u ( z ) , but may be not single-valued. Define

A 2 R

f ( z ) = (u(2) + i.(Z)) - - log 2,

357

then f ( z ) is a single-valued analytic function on (0 < Iz1 < 1) and Ref(z) is identically zero on { IzI = 1).

5 u,zn be the Laurent expansion of f ( z ) on (0 < IzI < Let f ( z ) = n=-m

00

l}, then lim = 0 and lim 5 1. Define g(z) = C bnzn, n=-m ,--too n-cc

- satisfying 6-, = -bn for n = 0,1,2, .. 0 , and b-, = a_ , for n = 1 , 2 , . . -. Then g ( z ) is an analytic function on (0 < IzI < +m}. When IzI = 1, it follows from Rebo = 0 and

-1 03 03

n=-cc n = l n = l n= l

00

that Reg(z) = 0. c,zn is an analytic function in

(1.1 < 1) and Re(f(z) - g(z ) ) is identically zero on (1.1 = 1). Consider P ( z ) = ef( ' ) -g( ' ) which is analytic and does not assume zero in { IzI < l}, and IF(z)I = 1 on (1.1 = l}, by the maximum and minimum modulus principles, we have F ( z ) i~ eia , hence f(z) = g ( z ) + ia.

Then f (z ) - g(z) = n=O

From the above discussion, we finally obtain

+cc A

b,zn + -log Iz], 21r

u ( z ) = Re n=-m

- where b-, = -bn and lim = 0.

n+cc

5123

(a) Let f ( z ) be a holomorphic function in the disc IzI 5 T whose zeros in this disc are given by all a2, , a , counted with multiplicity. Suppose further that luj I < T for all j = I, 2,. . . , n, and If(0)l = 1. Jensen's formula states that

Prove this.

(j = 1 ,2 , . +. , n) such that ( u j ( 5 t . Using Jensen's formula, show that (b) With the hypotheses and notations of (a), let n(t) be the number of uj

358

(c) For r < R deduce an estimate on n(r) in terms of max log If(Reie)l.

(d) What can be said about the zeros of bounded holomorphic functions in 0 < 8 5 2 x

the unit disc?

Solution. (a) Let

(Hamard)

then F ( z ) is holomorphic and has no zero in the disc ( 1 . ~ 1 5 r } , which implies that log IF(.)/ is harmonic in (1.1 5 r} . By the mean value theorem of harmonic functions,

(b) It is obvious that log 2 = h:j, $. By the definition of the function la3 I

n(t) we have

which shows that the identity holds. (c) Apply the identity in (b), we have

& Lzx log Jf(Re2')ldB = R

Denote max log If(Re")J by M ( R ) , we obtain

R n(r) 5 M(R)/log -.

0 < 8 < 2 R

359

(d) Let f (z) be a bounded holomorphic function in { z : IzI < 1). We know that f(z) can have countably many zeros. Suppose z = 0 is a zero of f (z ) of multiplicity m 2 0 with = a, and let the other zeros be ordered by 0 < lull 5 la21 I -... Obviously Ian1 -+ 1. Apply Jensen's formula in (a) to F ( z ) = with 0 < T < 1 such that there is no zero of f on ( 1 . ~ 1 = T } , we

Since f(z) is bounded, we assume

For any n, we can choose T such that T > lun[, and hence

Let T -+ I, we obtain n

00

which implies that the series C (1 - luj I) is convergent. j=1

SECTION 2 GEOMETRY OF ANALYTIC FUNCTIONS

5201

Find a one-to-one holomorphic map from the unit disk (1.1 < l} ont

(SUNY, Stony B slit disk {IwI < 1) - {[0,1)).

Solution. We construct the map by the following steps:

. z + 1 z - 1 21 = 41(z) = 2- : { z : IzI < 1) ---f (21 : Imzl < O};

{zi : Imzl < 0) ---t {z2 : 1.~21 < 1 and Imzz > O};

4 3 ( 2 2 ) = 22” : {z2 : 1221 < 1 and Imz2 > 0} +

{w : Iw( < l}\{w : Imw = 0,O 5 Rew < 1).

w 1

Then w = 43 0 4 2 o 41(z) = f ( z ) is a one-bone holomorphic map from the unit disk {IzI < l} onto the slit disk {IwI < 1}\{[0,1)}.

5202

(a) Find a function f that conformally maps the region { z : largzl < 1) one-to-one onto the region {w : 1wI < l}. Show that the function you have found satisfies the required conditions.

(b) Is it possible to require that f(1) = 0 and f(2) = $? If yes, give an explicit map; if No, explain why not.


(a) < = f l (z) = z: = , 5 logr (log 1 = 0) is a conformal map of { z : largzl < l} onto {< : Re( > 0}, and w = f 2 ( ( ) = 5 is a conformal map of {< : Re< > 0) onto { w : IwI < l}. Hence

361

is a conformal map of { z : largzl < 1) onto {w : IwI < 1) with f(1) = 0 and f (2) = 2q--1.

2r+1 -. ._ (b) Suppose W = f (z ) is an arbitrary conformal map of {z : largzl < 1) onto

{W : 161 < 1) with f(1) = 0. Then w = F(6) = foY-'(G) is a conformal map of { G : 161 < 1) onto {w : IwI < 1) with F ( 0 ) = 0, and W = F(w) = fof - ' (w) is a conformal map of {w : IwI < 1) onto {W : 161 < l} with F ( 0 ) = 0. By Schwarz's lemma, we have both [F(G) ( 5 161 and IF (w) l 5 Iw(, which implies that If(.)( = lY(z)I for every z E { z : (argzl < 1). Since

- - -

- we cannot require that f(2) = i.

5203

(1) Find one 1-1 onto conformal map f that sends the open quadrant ( ( 2 , ~ ) : 2 > 0 and y > 0) onto the open lower half disc {(z,y) : x 2 + y2 < 1 and y < 0).

(2) Find all such f . ( Toronto)

(1) Let C = 41(z) = z2 . It is a conformal map of { z = 2 + iy : 2 >

Let w = 4 2 ( ( ) = d m + ( , where d-1 = -ai. It is a conformal

Then w = 4 2 o &(z) = d n + z2 , where d-1 =; = -ai is a required conformal map.

(2) I f f is an arbitrary conformal map satisfying the condition of (l), then 4T1 o f o 4F1(C) is a conformal map of the upper half plane onto itself, which can be represented by $([) = s, where a, b, c, d E IR, ad - bc > 0. Hence f can be written as 4 2 o II, 0 &(z) .

Solution.

0 and y > 0) onto {C = < + iv : 77 > 0).

(=a

map of {C = < + iq : 7 > 0) onto {w = u + iw : u2 + v2 < 1 and w < 0).

z=e 4

5204

Map the disk { Izl < 1) with slits along the segments [a, 11, [-1, -b] (0 < a < 1,0 0. Compute f'(0) and the lengths of the arcs corresponding to the slits.

Solution. (Haruard)

We construct the conformal mapping by the following steps. ( i ) z 1 = 4 1 ( z ) = z + i : { I z I < 1 } \ { [ a , 1 ] U [ - 1 , - b ] } + ~ \ { [ - b - ~ , a + ~ ] } .

It has the point correspondences # q ( O ) = 00, 4l (a) = a + +, 41(b) = -b - i;, 1

#1(1) = 2 and &(-1) = -2. z i + ( b + ")

(ii) z2 = 4 2 ( 4 = -Zl+(aP+) the point correspondences

42(..) = -1,

and

4 2 (2)

and it is easy to know that

: @\{ [-6 - ; , a + i]} -+ C\[O, +a). It has

1 1 + b + -) < 0. b ( 4 2 0 41)'(0) = - (a + (iii) z3 = 43(~2) = fi : C\[O, +m) -+ (z3 : Imz3 > 0). It has the point

correspondences

and

For the convenience of computation, let

L - 4 &+4 &+&' 6

A = B = 2-4'

We also know that +i(-l) = -i. that 44(i) = 0 and d k ( i ) = -f.

(iv) w = 44(z3) = ~3 : {z3 : Imz3 > 0) --t {w : lwl < 1). It is obvious .rs+i

363

Now we define w = f (z) = 4 4 o 4 3 o 4 2 o &(z) . From the above discussion, we know that f maps the unit disk with slits [-1, -b] and [a, 11 conformally onto the unit disk with f(0) = 0 and

What correspond to the slits are the arc with endpoints and 2 and containing containing point z = -1 and the arc with endpoints

point z = 1. The lengths of the two arcs are

A - i A + i 5 4 A+z A - i T + &

I1 = arg- - arg- = 4arctgA = 4arctg

and B + i B - i 1 $4 B - 2 B + z B A + & - I2 = xg- - arg- = 4arctg- = 4arctg

5205

Let 0 < E < a, let denote the arc {eit : E 5 t 5 27r - E } and let fir be the complement of yE in the Riemann sphere. I f f is the conformal map of the unit disk onto a,, f(0) = 0, f'(0) > 0, describe the part of the unit disc that f maps onto (1.1 > I}.


We are going to find the map f by the following steps:

364

C . (shown in Fig.5.4), with q53(e-+a) = 0, arg$$(e-T*) = f + 5 , where D1 is a domain bounded by {C = eie, 5 B 5 27r - ;} and an circular arc I, which is orthogonal to c . = 1) and connects points e f* and e - T a in { / 0. After considering the boundary correspondence, we know that 1, corresponds to the arc { z = eit : It( < E } under the map @. Since the symmetric domain of (1.1 < 1) with respect to arc { z = eit : It( < E } is (1.1 > l}, and the symmetric domain of D1 with respect t o I, is D2 = (111 < l}\n1, by the reflection principle, @ ( z ) can be extended to a conformal map of fie onto {C : ICI < 1). Hence the conformal map f in the problem is nothing but the inverse of @, and the domain f maps onto (1.1 > 1) is D2, which is bounded by circular arcs I, and {C = eie : (01 5 i}.

5206

Suppose that w = f (z) maps a simply conncted region G one-to-one and conformally onto a circular disk DT with center w = 0, radius r , such that f ( a ) = 0 and If'(a)l = 1 for some point a E G.

(1) Prove that the radius r = r (G,a) of D, is uniquely determined by G and a.

(2) Determine r (G ,a ) if G is the region between the hyperbola zy = 1 (z > 0, y > 0) and the positive axes, and if a = 1 + i. Solution.

(1)Suppose c = g ( z ) is another conformal map of G onto a circular disk D,, with center C = 0 and radius T I , such that g(a) = 0 and Ig'(a)l = 1, then w = F ( C ) = f o g-'(C) is a conformal map of {C : 1" < r1) onto {w : Iwl < r }

(Indian u)

365

with F ( 0 ) = 0 and IF'(0)l = l # l = 1. Apply Schwarz's lemma to F(C) and we have IF'(0)l 5 2, hence TI 5 T . For the same reason, apply Schwarz's lemma to F- ' (w ) and we have T 5 TI , which implies TI = T . In other words, T is uniquely determined by G and a.

(2) We construct a conformal map of G onto a circular disk D, in the following steps:

z1 = & ( z ) = z2 : G + { q : 0 < Imzl < 2 } ,

with 41(1+ f) = $ + i, 11#:(1+ f)./ = A. 2 2 = 42(z1) = eHZ1 : (21 : o < Imzl < 2 ) -+ (22 : Imzz > 01,

with 42(: + i) = icy, Id;($ + i)l = f e y .

with 43(ie%") = 0, Irj5i(iegx)I = 2 d % l r e t = *

Define f ( z ) = 4 3 . 4 2 o 41(z), then w = f ( z ) : G ---f {w : JwI < -&}, with * f ( a ) = 0 , If'(a)I = 1. Hence r ( G , a ) =

5207

Let T ( z ) = 3 be a Mobius transformation. (a) Assume that z1, z2 E a' are two distinct fixed points for T , i.e., T ( z i ) =

zi, i = 1,2. Show that there exists a constant c such that

T ( z ) - z 1 z -21

T ( z ) - z2 = c-.

z - z2

(b) Use (a) to find an expression for T " ( z ) , n = 1,2 ,3 , .. -, if 1 - 32

2 - 3 T ( z ) = -.

(Iowa) Solution.

invariant under Mobius transformations, we have (a) Let a E a' be a point different from z1, z2. Because the cross ratio is

366

which is

Denoting

we obtain

T ( a ) - z1 T ( a ) - 3 2 . a - 3 2

T ( z ) - z 1 - z - z 1

a - z1 = c , .-

- c-. T ( z ) - ~ 2 z - z Z

(b) Since T " ( z ) = T(T"-l(z)), it is easy to have

z - z1 Cn -. ... = - - T"(3) - 31 - T"-l(z) - z1 = czT"-2(z) - 31

T"(z) - 2 2 CT"- l ( z ) - 2 2 T"-2(z) - 22 z - z2 -

When T ( z ) = s, by solving the equation = z, we obtain that z = f l are two fixed points of T. Choose a = 2, then T(a) = 5, hence c = 5-1 5+1 : 2-1 2+1 - - 2.

It follows from T"(z)- l 2 - 1 T n ( z ) + l z + l

= 2"-

that (2" + 1). - (2" - 1) (2" + 1) - (2n - 1)z'

T"(z) =

5208

(a) Justify the statement that "the curves

- 1 +-- 22 Y2 a2+X b2+X

form a family of confocal conics". (b) Prove that such confocal conics intersect orthogonally, if a t all. (c) Show that the transformation 20 = $(z+:) carries straight lines through

the origin and circles centered at the origin into a family of confocal conics.

Solution. (a) Without loss of generality, we assume a > b > 0. When -a2 < X < -b2,

the curves form a family of hyperbolas, while when X > -b2, the curves form

(Humurd)

367

a family of ellipses. Suppose the focuses of the conics are ( ~ c ( X ) , O ) . When -a2 < A < 4 2 ,

.(A) = J(a2 + A) + [-(b2 + A)] = m. When A > -b2 ,

Hence the curves 2 2 Y2

a 2 + A b 2 + A +-=1

form a family of confocal conics. (b) Suppose (xO, yo) is the intersection point of

= 1 L 1 : - + - X2 Y2 u 2 + A i b 2 + X i

and

Y2 - 1, L 2 : - + - - X2

a2+ A2 b 2 + A2

where A 1 # X 2 . It follow-from

and

that Y; 4 +

(b2 + Al)(b2 + A,) = O. (u2 + Al)(U2 + A,)

Noting that the tangent vector of L1 at (xo,yo) is t i = (&, *), and the tangent vector of L2 at (xo,yo) is 7 2 = ($&, &), we have

which implies that the confocal conics intersect orthogonally, if at all. (c) Let z = Tez', and

1 1 1 1 i 1 w = u + iv = - (z + -) = - (T + -) cos 8 + -(r 2 T - -) sin 8.

2 Z 2 T

The image of straight lines through the origin is

which are hyperbolas in w-plane. Because

cos2 e + sin2 e = I,

the focuses of the hyperbolas are ( f l , 0). The image of circles centered at the origin is

which are ellipses in w-plane. Because a(r + :)2 - $(r - :)' = 1, the focuses of the ellipses are (fl, 0). Hence the transformation

1 1 w = z ( z + ;)

carries straight lines through the origin and circles centered at the origin into a family of confocal conics.

5209

If f : D(0,l) = { z : IzJ < 1) --fa' is an analytic function which satisfies f (0) = 0, and if

IRef(z)l < 1 for all z E D(0, l),

prove that

Solution. It is easy to know that

is a conformal mapping of the domain {w : lwl < 1) with g(0) = 0. Hence

(Indiana)

e Y c + 1

{C : lReCl < 1) onto the unit disk w = F ( z ) = g o f (z ) is analytic in

369

D(0,l) and satisfies F ( 0 ) = 0 and IF(.)[ < 1. By Schwarz's lemma, we have IF'(0)l 5 1. Because

it follows from f (0) = 0 that

4 I f ' (0 ) l I -.

7r

5210

Let R = { z E a'; -1 < Imz < l}, and let F be the family of all analytic functions f : R -+a' such that I f 1 < 1 on R and f(0) = 0. Find

(Indiana) Solution.

is a conformal mapping of fl onto the unit disk with the origin fixed. For any analytic function w = f ( z ) : fl --+a' such that I f 1 < 1 and f (0 ) = 0, we consider the composite function w = F(C) = f o f{'(C). F(C) is analytic in the unit disk such that IF(C)I < 1 and F ( 0 ) = 0. By Schwarz's lemma,

IF(C>l I ICI. .E

Choose (0 = w, we have e T + 1

The equality holds if and only if F(C) = e"C, which implies

and the supremum is attained by f (z ) = e"fo(z), where 8 is a real number.

370

5211

Let f be an analytic function on D = { z ; IzI < 1) such that f ( 0 ) = -1, and suppose that 11 + f ( z ) I < 1 + I f ( z ) I whenever JzJ < 1. Prove that If’(0)l 5 4.

Solution.

1 + If(z)I that f (0 ) C R.

(Indiana)

Let R = C\{W = u + iv : u 2 0 and w = 0). It follows from 11 + f ( z ) I <

Set g(w) = E, (fit = 2) . Then g o f ( z ) is an analytic function w = - 1

on D with g o f ( 0 ) = 0 and 1g o f ( z ) I < 1. By Schwarz’s lemma,

I(s O f)’(O)l L 1.

we obtain If’(O>l L 4.

5212

Let P be the set of holomorphic function f on the open unit disc so that (i) Both the real and imaginary parts of f ( z ) are positive for IzI < 1, (ii) f ( 0 ) = 1 + i. Let E = {f(i) : f E P}. Describe E explicitly.


Let f E P and define

f2(z) - 2i < = F ( z ) = f2(z) + 2i’

Then F is a holomorphic function on the unit disc with F ( 0 ) = 0 and IF(z)I < 1. By Schwarz’s lemma, we have IF(z)I 5 121, which implies IF($)] 5 i. It should be noted that when f changes in P, F ( i ) can take any value in the disc {C : (that is the inverse of C = s) is a 23 1+c

1-c 5 f}. Because w =

37 1

conformal mapping of { C : the set {f2(i) : f E P} is equal to

5 :} onto {w : Iw - $'il 5 $}, we obtain that

1 0 . 8 lr 4 20 3 3 2 - 5 3 { w : I w - - z ~ < - } = { w = p e * + : I d - - 1 < a r c s i n - , p ~ - - p s i n d + 4 < 0 ) .

Hence

E = {f(-) : f E P } = {reio : 18 - - 1 < -arcsin - ,r4 - -r2sin28 + 4 5 0).

If we denote the two roots of p2 - y p s i n 4 + 4 = 0 by PI(+) , p2(4) where pl(q5) l}.

If F is the family of all analytic function on R such that I f 1 5 1 in R and lim f ( w ) = 0, find sup If(8)l. Your answer should be an explicit number,

f € 3 W + C C

and you should prove your assertion.

Solution.

map of D = { z : IzI < 1) onto R with d(0) = 00 and 4(4 - a) = 8.

and IF(z)I 5 1. By Schwarz's lemma,

(Indiana)

Define w = + ( z ) = 2(5+ :), it is easy to know that w = 4 ( z ) is a conformal

Then F ( z ) = f o 4 ( z ) = f (2( 5 + 4 ) ) is analytic in D and satisfies F ( 0 ) = 0

IF(.)I 5 1.4. Hence

If(8)l = IF(4 - &)I 5 4 - a. This upper bound can be reached if we let f = q5-l which belongs to family 7 and satisfies 4-l(8) = 4 - a. So we obtain

sup If(8)l = 4 - a. f E 3

372

5214

Let D be the upper-half and let f # id be a conformal map of D onto itself such that f o f = id. Prove that f has a unique fixed point inside D.

Solution.

where a, b, c, d E IR and ad - bc > 0. Then

(SUNY, S t o n y Brook)

Since f is a conformal map of D onto itself, it can be written as f ( z ) = s, (a2 + bc)z + b(a + d)

f ( z ) = C(U + d)z + d2 + bc *

It follows from f o f = id that b(a + d) = c(a + d) = 0 and a2 + bc = d2 + bc # 0.

If a + d # 0, then b = c = 0. Hence ad-bc > 0 and a2+bc = d2+bcimpies f = id, which contradicts the condition f # id. Thus we have a + d = 0 and the inequality a d - bc > 0 can be written as bc + a2 < 0.

= z , which is equivalent to cz2 + (d - a)z - b = 0. Since A = (d - a)2 + 4bc is equal to 4bc + 4a2 < 0, we know that f ( z ) = z has two conjugate roots, one in the upper-half plane and the other in the lower-half plane. So f has a unique fixed point inside D.

Now we consider the equation f ( z ) =

5215

Let fl be a convex, open subset of a‘ and let f : R ---f a‘ be an analytic function satisfying Ref’(z) > 0, z E 51. Prove that f is one-to-one in 51 (i.e., f is injective).

Solution.

is the line segment connecting z1 and 22. Since R is convex, L c il, we have

( I n d i a n a )

Let z1 # z2 be two arbitrary points in R. L : z ( t ) = zl +t (zz - z l ) , t E [0, 13

Hence

Since Ref‘(z) > 0 for z E Q, we know that Ji f‘(z(t))dt # 0, which implies f ( z 1 ) # f ( z 2 ) whenever z1 # 22.

373

5216

Show that if the polynomial P ( z ) = a,zn + un-1zn-' + + alz + ao, n > 1, is one-to-one in the unit disk IzI < 1 and a1 = 1, then [nu,[ 5 1.

Solution. It follows from the univalence of P ( z ) in { IzI < 1) that P'(z) = nu,^^-^ +

(n- l ) ~ , - l z " - ~ + . . . + 2 u ~ z + u l # 0 for all z E (1.1 < 1). In other words, the roots of P' ( z ) are all situated outside the open unit disk. Let z1, zz , . . , z,-1

be the roots of P' ( z ) , then Izj 2 1 for j = 1,2 , .. . , n - 1. Because P' ( z ) can also be written as nu, ( z - z1)(z - 2 2 ) ( z - z,-l), by comparing the constant terms, we have

(SUNY, Stony Brook)

n-1

(-l)n-lna, zj = a l . j=1

Since a1 = 1, we obtain

Inu,l = ~ l a l ' 5 1. n-1

j=1 n 14

5217

Let P ( z ) be a polynomial on the complex plane, not identically zero; let

(a) If all roots of P ( z ) lie in H , show that the same is true for the roots of

(b) For any non-vanishing polynomial P ( z ) , use the result in (a) to show

H = { z : Rez > 0).

dP/dz .

that the convex hull of the roots of P ( z ) contains the roots of d P / d z .

Solution. (Courant Inst.)

(a) Let z1, z2 , . . , z, be the zeros of P ( z ) . By assumption,

Rezj > 0 ( j = 1,2, . . . ,n) ,

and P ( z ) = u(z - zl)(z - z 2 ) . . . ( z - z,). It follows that

1 1 1 + - + ...+ - P ' k ) - - (logP(z))' = - - P ( z ) z - 21 z - 22 z - z,

374

When z E { z : Rez 5 0}, then < arg(z-zj) < $, or equivalently, Re& < 0. Hence Re C & < 0, which shows p(,) can not be zero on { z : Rez 5 0).

(b) Let z1, z2, . . . , z, be the zeros of P ( z ) , and I is a directed straight line passing through two zeros zk and zi such that the other zeros are on the right side of I (including on I ) . Denote the intersectional angle from the positive direction of the imaginary axis to I by 8. When z is on the left side of I, we have Re{e-ae(z - z j ) ) < 0. Hence

n

j=1

which shows that the zeros of P’(z ) do not lie on the left side of I . After considering all the directed straight lines passing through two of the zeros of P ( z ) such that the other zeros are on the right side of the line, we obtain that the zeros of P’(z) lie on the convex hull of the zeros of P ( z ) .

5218

Let f ( z ) be a Laurent series centered a t 0, convergent inC\{O}, with residue

(a) Show that there exists < on { z b at z = 0.

: IzI = 1) with

(b) Characterize those functions with


t o o

n=-cu (a) Let f (z ) = C bnzn, then

Hence

375

If If(() - <-l1 < J b - 1) holds for all C with I(I = 1, then

which is a contradiction. Hence there exists ( with = 1 such that

(b) If max I f ( < ) - ICI=1

= Ib - 11, it follows from

that If(C) - P I = Ib - 11

holds for all C with I ( I = 1.

real-valued function. It follows fiom Let f(() - <-l = ( b - l)ei+(e), where ( = eie and 4(8) is a continuous

that

which implies that (p(8) = -8, and hence

b - 1 f ( 0 - c-l = 7

holds on {C : 1[1 = 1). Apply the discreteness of zeros for analytic functions to f (z ) - p, we obtain f(z) = p, z EC\{O}.

5219

Assume f is analytic in a neighborhood of 0, f maps D into D , and f

(a) Show that Vz E aD, f’(z) # 0. (b) Show that -&[argf(e”)] > 0 for 8 in R.

maps d D into a l l , where D = { z : IzI < 1).

376

(c) Assume that f (0) = f'(0) = 0 and f l a ~ is a two-to-one map from a D

( Indiana-Purdue) onto dD. Show that f(z) # 0 whenever 0 < IzI < 1.

Solution. (a) Assume f'(z0) = 0, where zo E d o . Let f(z0) = wo E all . Then

where n 2 2, g ( z ) = b o + b l ( z - z o ) + b z ( z - ~ 0 ) 2 + . . . ,

with bo # 0. Let I' be an arc in defined by I' = { z E % : ) z - zoI = r ) , and denote by Ar#(Z) the change of # ( z ) when z goes along the arc I' in the counterclockwise sense. It is demanded that r is sufficiently small such that Ararg(z - zo) > when z E I?. It follows from f (z ) - wo = ( z - zo)"g(z) (n >_ 2) , that

and Ig(z) - bol <

3T n- 2 3

Ararg(f(z) - W O ) = nArarg(z - ZO) + Arargg(z) > - - - > K ,

which implies that f (z ) assumes values outside the disk when z E I'. It is a contradiction to the fact that f maps D into D. Hence f'(z) # 0 for all 2 € aD.

(b) Let z = rei6, and w = f(z) = Rei@. A variation of the Cauchy-Riemann equations for analytic function w = f(z) is

Since f maps a D into a l l , we know that %(ei6) = 0. If g ( e i e ) 1 0 , then at point ei6, = 2 = 2 = 0, which implies that g ( e i 6 ) = f'(ei0) = 0. But from (a) it is impossible. If g ( e i e ) < 0 , it follows from 7-s = Rg

aR i6 that %(e ) < 0. Since R = 1 when r = 1, g ( e i 6 ) < 0 implies that R > 1 when r < 1. This is also impossible. Hence we obtain

=

(c) Because f l a ~ is a tweto-one map from 8 0 onto aD, &Alzl=largf(z) = 2, which implies that f (z ) has two zeros (counted by multiplicity) in D. Since f(0) = f'(0) = 0, z = 0 is a zero of f of multiplicity m = 2. Hence f(z) has no zero in (0 < It1 < I}.

377

SECTION 3 COMPLEX INTEGRATION

5301

Evaluate the integral

(Indiana) Solution.

around z = 0 is: Function eez is analytic in { z : 0 < IzI < +co}, and its Laurent expansion

The coefficient of the term $ in the above development is

1 1 1 + 1 + 2! + . .. + ~ + . - . = e. (n - l)!

By the residue theorem, we obtain

5302

Evaluate

where y is the positively oriented circle { Jz1 = 1). (Indiana)

378

Solution. is analytic in { z : 0 < IzI 5 l}, and with z = 0 as a

pole. The Laurent expansion of & around z = 0 can be obtained as follows: It is obvious that

1 - 1 - - sin3 z ( z - k.23 + $ 2 5 - . . . I ~

1

Hence the coefficient of the term p in the above development is $. By the residue theorem, we have

- 2aiRes ( 3 , O ) 1 = ~ i . sin z

5303

For what value of a is the function

f ( z ) = LZ (; + 5) coszdz

single-valued?

Solution.

Laurent expansion around z = 0 is:

(Indiana)

Function F ( z ) = ( $ + 5 ) c o s z is analytic in { z : 0 < IzI < +a}, and its

1 F ( z ) = (;+;)cosz= (:+;) (1. 1 2 t 4 ! z 4 - . . . 1

= 2+('-2)P+(;-;)Z+.... U a 1

The necessary and sufficient condition for f (z ) to be single-valued is that in the the residue of F ( z ) a t z = 0 is zero, i.e., the coefficient of the term

above development is zero. Hence we obtain a = 2.

379

5304

Define h ( z ) = L m ( l + zte-t)-le-t cos(t2)dt.

What is the largest possible P so that h ( z ) is analytic for IzI < P?

Solution. ( Indiana-Purdue)

When z = -e,

dt . et - et

It is easy to see that when t -+ 1,

A (t - 1)2’

-- cos(t2) et - et

where A = $ cos 1, which implies that the integral is divergent. Hence P can not be larger than e .

For any r < e , let IzI 5 T . Consider the integral

dt. et + z t

It follows from let + ztl 2 et - rt and the convergence of the integral

dt

that

is uniformly convergent in any compact subset of { z : IzI < e}. By Weierstrass theorem, we know that h ( z ) is analytic in { z : IzI < e} . Hence the largest possible P is equal to e.

5305

Let f ( z ) be analytic in S = { z €67; IzI < 2). Show that

380

(Iowa) Solution.


It follows from the above three equalities that

5300

Suppose that the real-valued function u is harmonic in the disk {IzI < 2}, 'u is its harmonic conjugate and u(0) = v(0) = 0. Show that

d z d z z

where y(t) = eZnit , t E [0, 11.

Solution. (SUNY, Stony Brook)

Let f ( z ) = u ( z ) + iw(z). Then f (z ) is analytic in { z : IzI < 2}, and we have

38 1

It follows from

and

that

f (z ) +m u ( z ) =

f (.> - f w(z) =

2 -

22

5307

Let f be an analytic fqnction on an open set containing D(0,l) = { z ; IzI 5

(a) Prove that 11.

d”f(0) = /2T e-nie[Re f (eie)]dO. dzn 0

(b) If f(0) = 1, and if Ref(z) > 0 for all points z E D(0, l), prove that

(Indian u)

382

Solution. (a) Assume that

00

k=O

we have

By Cauchy Integral Formula,

Hence

(b) Because Ref(z) is harmonic on D(0, l), by the mean-value formula of harmonic functions,

Ref(e")de = Ref(0) = 1. k I'" Noting that Ref(eis) 2 0, we have

= 2(n!) .

383

5308

I f f is analytic in the unit disk and its derivative satisfies

IfWI I (1 - 14)-1, show that the coefficients in the expansion

n =O

satisfy lunl < e for n 2 1, where e is the base of natural logarithms.


00 It follows from

f ( z > = C anzn n=O

00 that

f’(2) = c nunzn-l, n = l

where nu, = - 1 f ) o d z , (0 < T < 1).

It is obvious that lull = If’(0)l I 1 < e .

For n > 1, we choose T = 1 - :,

1 (1 + -)-I < e. n - 1

5309

Let f = u + iw be an entire function.

384

(a) Show that if u 2 ( z ) 2 wz(z) for all z EC, then f must be a constant. (b) Show that if If(z)I 5 A+BIzlh for all z € @ w i t h some positive numbers

A, B , h, then f ( z ) is a polynomial of degree bounded by h.

Solution. (a) Let

( Stanford)

F ( z ) = e - f 2 ( z ) = , - (ua(z)-v2(z))-Ziu(L)v(r)

Then F ( z ) is an entire function with

-(.'(.)-."(.)) 5 1. IF(z>I = e

By Liouville's theorem, F ( z ) must be a constant, which implies that f ( z ) is a constant.

(b) Let

For any integer n > h,

A + B R ~ R" . 5

Letting R -+ +m, we obtain that a, = 0, which implies that f ( z ) is a polynomial of degree bounded by h.

5310

Let f be an entire function that satisfies IRe{f(z)}I 5 lzln for all z, where n is a positive integer. Show that f is a polynomial of degree at most n.

Solution.

theorem that when IzI < R,

(Indiana)

Let R be an arbitrary positive number. Then it follows from Schwarz's

385

Especially when IzI = +, 1

If(.)l L -3R" * 2n + IIm{f(O))l = 3R" + IIm{f(O))lr

which implies that there exist constants A , B such that

If(z>I 5 AlzY + B

holds for all z EC. Let

where

Hence when k > n,

which shows that f(z) is a polynomial of degree at most n.

5311

Compute the double integral

1 J , cos z d x d y

where D is the disk given by { z = x + i y E a' : x 2 + y2 < 1).

Solution. (Iowa)

First we have the following complex forms of Green's formula:

386

The problem can be solved directly by either one of the above two forms:

or

1 sin zd? = - - sin zd( -)

JJDcoszdxdy = Z

sin z -dz = R.

5312

n = O

be analytic in D = (1.1 < 1) and assume that the integral

is finite. (a) Express A in terms of the coefficients an. (b) Prove that

for z E D. (Indiana)

Solution. (a) BY

n = l

we have

387

Noting that

we obtain that

.1 00 00

(b) By Cauchy’s inequality, we have

5313

Let f be analytic in (0 < IzI < 1) and in L2 with respect to planar Lebesque measure. Is 0 a removable singularity? Proof or counterexample.

Solution. ( Stanford)

The answer to the problem is Yes. Let the Laurent expansion of f in {z : 0 < IzI < l} be

00

From

388

we have

Let E < 1 be a small positive number, and then

1' la, 12r2n+1dr < - 21r

Then a, must be zero when n 5 -1. Otherwise, let E -+ +0, the left side of the above inequality will tend to infinity, while the right side of the inequality is finite, which leads to a contradiction. Hence

n=O

which shows that z = 0 is a removable singularity of f .

5314

Evaluate the integral

(Indiana) Solution.

Let z = peie, a = rei+.

When r < p,

When r > p,

5315

Evaluate

by the method of residues. (Columbia)

Solution. Denote

It is obvious that I ( u ) is an analytic function in { u : la1 > 1). Then we have

lr

2dx I ( a ) =

dx dx

390

Let z = eisl then

dz iz ' - dx =

z + 2-1 2 '

cosx =

and i d z

= iz,=l 22 - 2(2u + l )z + 1'

= iz,=l ( z - z1)(z - z2) z1 - z2

Denote the two roots of z2 - 2(2u + 1). + 1 = 0 by z1 and z2. Since z1 -z2 = 1, we may assume that lzll > 1, 1221 < 1. By the residue theorem we have

27T -- idz -

7T - 27r - - d(Zl -t 2 ~ ) ~ - 42122 - 2 d m ' is also analytic in { u : la1 > l}, and the

2%& It should be noted that

* ~ ,

branch of d m should be chosen by a r g d w 0.

5316

Consider the function

(a) Use the residue theorem to find an explicit formula for

2n

f ( z ) = 1 g(z,6)de

when ( z ( < 1. (b) Integrate the Taylor expansion

n=O

term by term to find the coefficients in the Taylor expansion

391

(c) Verify directly that (a) and (b) agree when IzI < 1. (Courant Inst.)

Solution. (a) Let c = ele. Then

C2 - 1 22 2ic '

,i0 - e - i O - sin8 = -

and

where ci = $(d=--l), (2 = i ( - d n - l ) , and the single-valued branch of d m in (1.1 < 1) is defined by 4- I r = ~ = 1. Because IC1 . C21 = 1,

(b) It follows from I sin81 5 1 and IzI < 1 that

00

g(z,e) = C ( - 1 ) ' s i n k 8 . z k , (1.1 < 1). k=O

Since the series converges uniformly for all 8 E [0,27r], the integration with respect to 8 can be taken term by term, and

f(z)=$' ( g ( - l ) k s i n k 8 . z k 00

k=O

where 2 r

Uk = (-1)' sink ode.

It is easy to obtain that uzn-l = 0 and

2R

(c) In order to verify that (a) and (b) agree when IzI < 1, we develop the function f(z) in (a) into a power series:

00 2-K f ( z ) = ~ - 27r(l- z"-i = 27r-y- l )"CIf ,Z2" d C - 7 - n=O

392

we know that the results in (a) and (b) agree when J z J < 1.

5317

If a is real, show that

exists and is independent of a. (UC, Irvine)

Solution. First we have

It follows from the existence of R hllRe-x2dz

that

exists.

r3 u r4 as shown in Fig.5.5. Define f (z ) = e-" and choose the contour of integration I' = rl U I'2 U

Fig.5.5

As f (z ) is analytic inside I', by Cauchy integral theorem,

393

dY - ie-R2 J,' ey2+2Ryi

= 0.

Letting R -+ 00, it follows from the facts that e-R2 -+ 0 ( R + CQ) and

5318

Let n 2 2 be an integer. Compute

Solution. (Iowa)

0 R

Fig .5.6

Let f ( z ) = &, and select the integral contour r as shown in Fig.5.6. f ( z ) has one simple pole z = eEi inside I'. By the residue theorem, we have

f ( z ) d z = 2 ~ i R e s ( f , e : * ) .

394

and

5319

Evaluate

with full justification.

Solution.

cos (x ') dx JD- (Minnesota)

Fig.5.7

Define f(z) = e - z 2 ,

3 and choose the contour of integration r = r j as shown in Fig.5.7. B

j=1 caus

f (z ) = e - I Z is analytic on I' and inside I', by Cauchy integral theorem, we have

3

j = 1

395

For the integral of f (z) on r2, we make a change of variable by w = z2, then

where 7F

7 2 = { w : lwl= R2,O 5 argw 5 -}. 2

By Jordan's lemma, we have

For the integral of f (z ) on r3, we have

It is well known that

f i l, f(z)dz = I" e-x2dz + - 2

when R -+ 00. Hence we obtain by letting R ---f co that

& 0 2 '

00

L m ( c o s z2 + sin x2)dz + i / (cos z2 - sin z2)dz = -

which implies

5320

Evaluate

(Iowa)

396

Solution.

Fig.5.8

Define

and select the integral contour inside r, by Cauchy integral theorem,

as shown in Fig.5.8. Because f(z) is analytic

where


lim I' iReie f (Reie)d6 = 0 R-r 00

and

397

Since the Laurent expansion of f about z = 0 is 00

where a-1 = -2i, we know that Res(f, 0) = -22. Letting E -+ 0 and R + ca, we obtain

7r dx = -.

2

5321

Let f (z ) be holomorphic in the unit disk IzI 5 1. Prove that

where respective integration goes along the straight line from 0 to 1 and along the positively oriented unit circle starting from the point z = 1. The branch of log is chosen to be real for positive z.

Solution. (SUNY, Stony Brook)

Fig.5.9

Let the contour of integration I' be shown as in Fig.5.9, and the single- valued branch of logz be chosen by argzl,=-I = T . Since f(z) logz is holomorphic inside the contour I?, by Cauchy integral theorem,

J, f ( z ) logzdz = 0,

398

where


Letting E + 0, we obtain

where the integration contour IzI = 1 has starting point and end point z = 1, and the value of logz at the starting point z = 1 is defined as 0.

5322

where a and b are complex constants, not both equal to zero.

Solution. First we assume la1 > lbl, and then the multi-valued analytic function

log(a + bz) has single-valued branch on { z : IzI 5 1). Take ei@ = z , then d4 = g, and

(Hamud)

2 r I log la + be'@]& = Re

= R e ( 2 ~ l o g a ) = 2rlog lal.

399

When la1 < lbl, we have

In the case la1 = lbl, let b = ae". Then

Fig.5.10

In order to evaluate the integral

we define log( 1 + z )

7 z f(.) =

where the single-valued branch is defined by log(1 + z ) lz=o= 0. Choose a contour of integration = r, Uy, as shown in Fig.5.10. Since f ( z ) is analytic on I' and inside r, by Cauchy integral theorem, s, f(z)dz = 0. Because

we have

I* log 11 + ei@1d4 = Re IR log(1 + e*+)d4 J -R J -R

400

Hence we obtain

log la + beiGldq5 = 2 7 ~ max(l0g la[, log lbl}. J,'"

5323

Evaluate

(Iowa) Solution.

Fig.5.11

Let

and select the integral path I? as shown in Fig.5.11. The single-valued branch of logz is chosen by argzI,,-1 = ?r. By the residue theorem, we have

J , f(z)dz = 2?riRes(f, -l),

401

where

(logx + 2ai)' 2 X

iRege f (Reie)d6 + dx

rO

+ 1; ice"f (&eie)d6.

It is obvious that

lim J,'" iReief(Reie)d6 = 0 R-rm

and

In order to find Res(f,-l), we consider the Laurent expansion of f about z = -1:

= c a n ( z + lyl n=-3

where a-1 = 1 - ai. Hence

2aiRes( f , -1) = 27ri + 2 2 .

AS E -+ 0 and R -+ 00, it turns out that

-4ai log x + 4a2 dx = 2ai + 2 2 .

Comparing the imaginary parts on the two sides of the above identity, we obtain

402

5324

Evaluate the following integrals: at (the integration is over the imaginary axis), +im

(a) S-im ( t 2 - 4 ) l o g ( r + l )

(b)

Solution.

&dz for CY in the range -1 < QI < 2. (Courant Inst.)

(a)

Fig.5.12

1 Define

The single-valued branch for log(z + 1) is chosen by log(z + l ) l z = o = 0, and the contour I' of integration is shown in Fig.5.12. As f ( z ) is analytic on and inside J? except a simple pole a t z = 2, we have

f ( z ) d z = 2 ~ i R e s ( f , 2),

where

and

Because

403

and

by letting E -+ 0 and R --+ 00, we obtain

dz Ti 2 +ice

= -(1- -). ,/,, ( z 2 - 4)log(z + 1) 4 log3

Fig.5.13

Define

The single-valued branch for zm is chosen by argzIr=z,O = 0, and the contour I.' of integration is shown in Fig.5.13. As f ( z ) is analytic on and inside r except a simple pole at z = e:', we have

f ( z ) d z = 2~iRes(f , e:')), I F

where

and

Res(f,efa) = lim ( z - efa) f ( z ) z+e: '

3e +a

e f i

1

- - -

-- - - 3e f ( 2 - m ) i '

404

Because 2.z

lirn f (Re ie ) iRe ied8 = 0 R-CU

2a when a < 2 and - lilil ' f (&eie) iseied8 = 0

when a > -1, by letting E -+ 0 and R + co, we obtain

7r d x =

5325

00 X u 7r( 1 - a ) Show that

J, ( 1 + x 2 ) 2 d x = 4 cos( y ) ' for -1 < a < 3, a # 1. What happens if a = l?

Solution. Let

where (argza)z=z>O = 0, and select the integral path I' as shown in Fig.5.14. By the residue theorem, we have

f ( z ) d z = 2aiRes(f(z), i ) ,

Fig. 5.14

where

405

+ J," iEeie f (&eie)d8

and

Res(f(z),i) = lim [ ( ~ ta 1'-9 - e i" 2 . 2'2 2 + i ) 2

It follows from a < 3 that

lim iReie f (Re")dB = 0 , R-CO

and from a > -1 that

Letting E + 0 and R --$ 00, we obtain

XQ a ( 1 - a ) 1" e 2 .

2

When a # 1,

a(1 - a ) - 2" a(l - a ) I- ( 1 + x y d X = 4 cos( y) '

when a = 1. T ( 1 - a ) 1 - X

5326

(a) Prove that

converges if 0 < a < 1

406

(b) Use complex integration to show that

?fa x P a cos x d x = sin - . I?( -a + 1).

2

(Hamad) S o h tion.

(4 03 1 03 1 e i x x - u d x = ( J! x - ~ cos x d x + x P a cos x d x ) + i x - ~ sin x d x .

It follows from a < 1 that

z - ~ cos x d x

is convergent. It is also obvious that

lLA cos x d x 1 5 2 ,

xPa is monotonic decreasing and

lim x - ~ = O X-++03

s i n x d x 5 2, LA I for (Y > 0.

By Dirichlet’s criterion, we know that s;” 2 - O cos x d x and also convergent. Hence

x - ~ sin x d x are e i z x - a d x is convergent when 0 < a < 1.

(b)

zRD iC o c R

Fig.5.15

Let f (z) = z - a e - z , and the contour of integration r is chosen as shown in Fig.5.15. The single-valued branch of f (z ) on I? is definde by z-aIz=2>o > 0.

407

By Cauchy integral theorem,

0 +L i&e"f(&eie)de = 0.

It follows from -a < 1 that

0 l i i L ice'e f (&eie)d8 = 0,

and from -a > 0 and Jordan's lemma that .* ilm 1 a iReie f (Reie)de = 0.

Letting E -+ 0 and R -+ 00, we have

Multiplying both sides by e y i , and comparing the imaginary parts, we obtain

T-a x - ~ cos xdx = sin -I?( --a + 1). I" 2

5327

Use a change of contour to show that

dt 7

provided that -a and p are positive. Define the left side as a limit of proper integral and show that the limit exists.


Since

408

is monotonic with respect to x and

= 0, lim - 1

x - + + a , x + p the convergence of the integral

dx

follows from Dirichlet's criterion. Define

and choose the contour of integration I' = I'l U I'2 U I'3 as shown in Fig.5.16.

Fig.5.16

By Cauchy integral theorem, we have

R --ax R - a x i (' - + l2 f ( z ) d z - 1 b d x = 0. = 1 x 2 + p z

It follows from Jordan's lemma that r

Letting R --+ 00 and considering the real part in the above identity, we obtain

409

5328

(a) Let c be the unit circle in the complex plane, and let f be a continuous a'-valued function on c . Show that

is a holomorphic function of z in the interior of the unit disk.

associated function F is identically zero.

Solution.

Choosing 6 > 0 such that 6 < p, we prove that

(b) Find a continuous f on c which is not identically zero, but so that the

(Minnesota)

(a) Let zo be an arbitrary point in the unit disk. Then 1 - lzol = p > 0.

has a power series expansion in { Iz - zo I 5 6). It is clear that

when Iz - z0I 5 6 and C E c. We can also assume I f (<)I 5 M because f is continuous on c . Thus

f (0 1 < - zo 1 - z--lo -. - f o = f (0 -

(C - 20) - ( z - zo) c - z C - 2 0

As

00

and (f)n is convergent, the series C (z)" converges uni-

formly for all E c . Hence termwise integration is permissible, and we obtain n = O n = O

410

Since zo is arbitrarily chosen in the unit disk, Ff ( z ) is holomorphic in { IzI < 1). (b) Take f ( C ) = f (Icl = 1). Then

1 1 1 F f ( z ) = / -d< = (- - t ) dC = ;(2ai - 2ai) = 0.

C

In fact, f (C) can be taken as - for any positive integer n and fixed zo E { z : IzI < 1). When c E c ,

5329

Let [a, b] be a finite interval in B and define, for z in D = C - [a, bll

f (z) = J." A. t - z

Show that f(z) is analytic in D. Given c, a < c < b, calculate the limit of f ( z ) as z tends to c from the upper half plane and as z tends to c from the lower half plane.

(UC, Iv ine) Solution.

For any zo E D , choose 6 > 0 sufficiently small such that { z : Iz - 201 5 6 } n { z = a : + i y : y = O , a L z I , b ) = 8 . Whenlz-zol < b , a < t < b , w e h a v e

00 1 1 -. - - 1 - 1

-- t - z ( t -zo) - (z -zo) t - - 0 1--

t - % a n = O

and the series converges uniformly for t with a I, t 5 b. Hence

41 1

holds for z E { z : Iz -201 < b } , which implies f(z) is analytic in { z : )z - zoJ < 6). Since zo is an arbitrary point in D, we obtain that f(z) is analytic in D.

For z E D, f ( z ) can also be represented explicitly by

where the single-valued branch is defined by arg r = s o > b = 0. Let rl and r2 be two continuous curves connecting z = xo > b and z = c in the upper half plane and the lower half plane respectively. Then the limit of f (z ) as z tends to c from the upper half plane is

a - b c - b log - I:::[ +iAr,arg- -

while the limit of f ( z ) as z tends to c from the lower half plane is

- log 1-1 + *i, c - - a z - - a

5330

For each z E U = { z : Imz > 0) define

1 sin2 t g(z ) = C l , t-a dt.

Determine which points a E B have the following property: there exist E > 0 and an analytic function f on D(u,E) such that f(z) = g ( z ) for all z E U n D(-a, E l .

(Indiana) Solution.

defined from point z = -1 to point z = 1, and define a function Let I' be the half unit circle in the lower half plane whose direction is

It follows from the Cauchy integral theorem that when z E U, f ( z ) 5 g(z) . With a similar reason as in problem 5328, f ( z ) is analytic in the complement of I?. Hence we obtain that for any a E a, a # fl , there exists E > 0

412

( E < min{la-ll, la+ll}) such that f (z ) is analyticin D(u,E) = { z : Iz-al < E }

and f (z ) = g ( z ) for all z E U n D(u,E). When a = 3 ~ 1 , such a f(z) does not exist. The reason is as follows: As

sin2 t sin2 t - sin2 z sin2 z +- t - z t - z t - z '

where sin2 t--sina -x is an analytic function of two variables for ( t , z ) E a' x a', we know that

2 ~ i 'J -' t - z

- --

t--x

dt ' sin2 t - sin2 z

h ( z ) = -

is analytic for z E C. But

sin2z ' sin2z z - 1 dlog(t - z ) = - dt = -Il 2Ti 2Ti log z+l'

which has branch points z = f l , hence g ( z ) can not be analytically continued to D(&l,&).

413

SECTION 4 THE MAXIMUM MODULUS AND

ARGUMENT PRINCIPLES

5401

Let u €67, JuJ 5 1, and consider the polynomial -

U a 2 2 2 P ( z ) = - + (1 - I.(">. - -z .

Show that ( P ( z ) ( 5 1 whenever 12) 5 1.

Solution.

- U a 2 2 2

z [ ( l - la12) + -(- - izz)].

~ ( z ) = - + (1 - IaIz)z - --z

l a 2 2

=

When IzI = 1,

U a - u a Re( - - Zz) = Re[- - (Cz)] = Re[- - -1 = 0,

IIm(-- - az)I _< 21al.

z Z z z a 2

Hence when ( z ( = 1,

l a 2 2

lp(z)12 = (1 - Ia12)>" + (Im[-( - - ~ i z ) ] ) ~

<_ (1 - 2[u(' + la14) + [aI2 = 1 - luI2 + laI4 5 I.

By the maximum modulus principle, IP(z)( 5 1 whenever IzJ 5 1.

5402

(Indiana)

Let f be holomorphic in the unit disk (1.1 < l}, continuous in (1.1 5 1) and If(.)/ = 1 whenever (zJ = 1. Prove that f is a rational function.

(SUNY, Stony Brook)

414

Solution. If f (z ) has infinite many zeros, by the isolatedness of the zeros of holomor-

phic functions, the zeros must have limit points on the boundary of the unit disk. But it will violate the fact that f is continuous in ( 1 . ~ 1 5 1) and I f ( . ) / = 1 whenever IzI = 1. Hence f has only finite zeros in the unit disk. Denote all these zeros by zlr 22,. . . , z,, multiple zeros being repeated, and define

Then F ( z ) is holomorphic in (1.1 < 1). continuous in (1.1 5 1) and IF(z)I = 1 when IzJ = 1. By the maximum modulus principle, IF(z)I 5 1 in ( 1 . 1 5 1). Since F ( z ) has no zero in (IzI 5 1) -L- is also holomorphic in (1.1 < l},

F ( z )

continuous in ( 1 . ~ 1 5 1) and I&JI = 1 when IzI = 1. Application of the maximum modulus principle to & yields IF(z)I 2 1 in (1.1 5 1). Hence IF(z)I = 1 holds in ( 1 . ~ 1 5 l}, which implies F ( z ) = eia with Q: a real number. So we obtain

5403

Let f be a continuous function on v = { z : IzI 5 1) such that f is analytic in U. If f = 1 on the half-circle y = (eie : 0 5 8 5 a}, prove that f = 1 everywhere in v. Solution.

Define F ( z ) = ( f (z) - l ) ( f ( -z ) - l), then F ( z ) is also continuous on V and analytic in U. When z E d U , we have either f (z) - 1 = 0 or f(-z) - 1 = 0. Hence F ( z ) = 0 holds for all z E v, which implies either f (z ) - 1 0 or f ( -z ) - 1 0, we obtain f ( z )

Remark. The condition that “f = 1 on the half-circle y” can be weakened to that “f = 1 on an arc y = (eie : 0 5 8 5 f } , where n is a natural number”. In this case, the proof is the same except that F ( z ) is defined by

(Indiana)

0. Since f(z) - 1 0 is equivalent to f(-z) - 1 1 for all z E V .

2n- 1 ~ ( z ) = ( f (z> - l)(j(ze:z) - l ) ( f ( ze+ i ) - ~ ) . . . ( f ( z e ? ~ a ) - 1).

41 5

5404

Let S denote the sector in the complex plane given by S = { z : -5 < argz < $}. Let 3 denote the closure of S. Let f be a continuous complex function on 3 which is holomorphic in S. Suppose further

(1) If(z)I 5 1 for all z in the boundary of S; (2) ~ f ( z + i y ) ~ 5 efi for all z + iy E S.

Prove that If(z)J 5 1 for all z E S.

Solution. Let F ( z ) = e- '"f (z ) , where E > 0 is an arbitrary fixed number. Then

F ( z ) is also continuous on 3 and analytic in S. When z is on the boundary of S, JF(z ) I = e-'"If(z)I 5 1. When IzI -+ $00 (-4 < argz < %), IF(z)I 5 e-€" . e f i + 0. By the maximum modulus principle, we have IF(z)I 5 1 for all z E S, which implies If(z)I 5 lea'[ = ear for all z E S. Because E > 0 can be arbitrarily chosen, letting E -+ 0, we obtain lf(z)I 5 1 for all z E S.

(SVNY, Stony Brook)

5405

Let K be a compact, connected subset of@ containing more than one point and let f be a one-to-one conformal map of@\K onto A = {z : JzJ < 1) with f(00) = 0. If p is a polynomial of degree n for which Ip(z)I 5 1 for z E K, prove that

M z ) I 5 If(z)I-" for z EG\K.

(Indiana) Solution.

Because f is a one-to-one conformal map of@\K onto A with f(w) = 0, it has a simple zero at z = 00. Since p is a polynomial of degree n, it has a pole of order n at z = M. Hence the function F ( z ) = p(z)fn(z) is analytic in G\K which contains point z = 00. As f ( z ) maps@\K onto A = { z : IzJ < l}, we have lim lf(z)J = 1. Together with Ip(z)I 5 1 for z E K, we know that the

limit of IF(.)\ when z tends to K can not be larger than 1. Apply the maximum modulus principle to F ( z ) on@\K, we obtain IF(z)I 5 1 for z EG\K, which implies Ip(z)I 5 If(z)I-" for all z E@\K.

%-+K

416

5406

Suppose f and g (non-constant functions) are analytic in a region G and of the region. Assume that E is compact. Prove continuous on the closure

that I f 1 + (g ( achieves its maximum value on the boundary of G.

Solution.

prove that if zo E G , then f and g must be constants.

(Iowa)

Assume that I f 1 + 191 achieves its maximum value c ( c > 0) at zo E z, we

Let I f(zo)I = f (zo)e*+’ , Ig(zo)I = g(zo)ei+2.

Then for fixed 41 and 4 2 ,

~ ( z ) = f (z)ei+l + g(z)e”Z

is analytic in G and continuous on G. It follows from

IF(.)I 5 If(.>I + Is(z)I L c ,

~ ( 2 0 ) = f (zo)e i+ l+ g(zO)ei” = ~f(zo)I + Ig(zo)I = c

and zo E G that ~ ( z ) = f (z )e i+ l + g(z)ei+2

must be the constant c. Without loss of generality, we assume that f is not a constant, and try to

lead to a contradiction. Since the image of an open set { z : Iz - 201 < 6) c G under f is an open set which contains point f(zo), f (z) assumes all the values f(z) = f(z0) + &ei+ for small E > 0 and 0 5 4 < 27r in { z : Iz - z0I < 6). Then when 4 + 41 # 0, ?r, we have

lf(.)I + lg(z)l = lf(.)I + Ic - f ( z ) e i 9 = l f (z0) + + ~c - f(zo)ea+l - &ei(+++l)l

- - l,cei(++h) + f(zo)e”l I + l&ei(+++l) - g(zo)ei@21

> f(zo)ei+l + g(zO)ei+Z = c,

which contradicts that zo is a maximum value point of I f 1 + lgl. Hence f must be a constant, which also implies g is a constant too.

417

5407

Suppose f(z) is an entire function with

Show that f(z) is identically 0. (Iowa)

Solution. For any R > 0, consider function

g ( z ) = (Z - Ri>(z + Ri) f ( z ) .

When IzI = R, and Imz 2 0, denote by 0 the angle between the line perpendicular to the imaginary axis and the line passing through z and Ri. Then 0 5 0 5 5, and

When Iz( = R, and Imz < 0, denote by 8 the angle between the line perpendicular to the imaginary axis and the line passing through z and -Ri. Then 0 5 0 < t , and

It follows from the above discussion that when JzI = R,

By the maximum modulus principle, when IzI < R,

Now fixing z, and letting R -+ +m, we obtain f(z) = 0. Since R can be arbitrarily large, we have f(z) = 0 for all t EC.

418

5408

Suppose f is analytic on (2; 0 < IzI < 1) and

Show that f E 0. (Indiana)

Solution. Denote the Laurent expansion o f f on { z ; 0 < IzI < 1) by

where

It follows that

When n < 0, letting r + 0, we have

a, = O (n= -1,-2,...),

which implies z = 0 is a removable singularity of f . In other words, f can be extended to an analytic function of the unit disk.

= 0 when IzI = 1. By the maximum modulus principle, we obtain

Since log 121

f 0.

5409

Let f be an analytic function on D = { z : IzI < l}, f ( D ) E D and f(0) = 0. (a) Prove that If(.) + f ( -z)I 5 2)zI2 for all z in D and if equality occurs

for some non-zero z in D, then f(z) = eiaz2.

419

(b) Prove that

(Indiana) Solution.

(a) Let F ( z ) = f(z) + f(-z), then F(O) = 0,

Hence is analytic in D , and when z tends to dD, the limit of can not be larger than 2. By the maximum modulus principle, If(.) + f(--z)I 5 21zI2 holds for all z E D .

I I If equality occurs for some non-zero z in D, we have

f(z) + f ( - z ) = 2eioz2,

where a. is a real constant. Let

M

n = l

it follows from f ( z > + f ( - z ) = 2eiaz2

that a2 = eio, a4 = a6 = ... = 0.

Because If(z)I < 1 for z E D , we have

Since a2 = e ia , the other coefficients must be zero, which implies f (z) = e iaz2 .

(b)

420

5410

I f f is analytic and lf(z)I < 1 on { z : IzI 5 l}, prove that f ( z ) has a fixed

(Rutgers) point.

Solution. Let F ( z ) = f (z) - z and G ( z ) = -z. When IzI = 1,

IF(z) - G ( z ) / = lf(z)I < 1 = IG(z)l.

By RouchC’s theorem) F ( z ) and G(z) have the same number of zeros in { z : IzI < 1). Since G(z ) has only one simple zero in { z : IzI < 1)) we conclude that f ( z ) - z has one zero in { z : IzI < l}, which implies that f ( z ) has a fixed point in { z : IzI < 1).

5411

Let f ( z ) = z+e- ’ , X > 1. Prove or disprove: f (z) takes the value X exactly once in the right half-plane. If the answer is yes, is the point necessarily real? Justify.

Solution.

curve I’ on the right half-plane, where

(Iowa)

Let R be a sufficiently large real number such that R > 2A. Take a closed

?r ?r I’ = { z = t + i y : t = 0, -R 5 y 5 R} u { z : Izl = R, -- < argz 5 -}. 2 - 2

Define F ( z ) = X - z - e-”

and G ( z ) = X - Z .

When z E I?, IF(z) - G(z)l = le-”l 5 1 < IG(z)I.

Since G ( z ) has exactly one zero inside I’, it follows from RouchC’s theorem that F ( z ) has exactly one zero inside I?. Because R can be arbitrarily large, F ( z ) has exactly one zero in the right half-plane. Hence f(z) takes value X exactly once in the right half plane.

42 1

Take z = x 2 0. We have

F ( x ) = X - x - e-",

which is a real-valued function of real variable x. Since F ( x ) is continuous and

lim F ( z ) = -m,

there must exist 20, 0 < xo < $00, such that F(z0 ) = 0. In other words, the point z in the right half-plane such that f ( z ) = X is necessarily real.

F(O) > 0,

x-++w

5412

Suppose f is analytic in a region which contains the closed unit disc {z : (z( 5 1). Assume f is non-zero on the unit circle {z : IzI = 1). Let C denote the unit circle traversed in the counterclockwise sense. Suppose that

and

Find the location of the zeros o f f in the open unit disc {z : IzI < 1).

Solution.

being repeated. Then

(Iowa)

Assume z1, z2, +. - , z, are the zeros of f ( z ) in {z : IzI < l), multiple zeros

n

f(z) = !dz) nc. - Zj), j=1

where g(z) is analytic and has no zero in {z : Izl 5 1). We have

422

It follows from (1) that n = 2, i.e., f ( z ) has two zeros in the unit disk. Then for f (z) = ( z - z1)(z - zz )g (z ) ,

and

which show that z1,2 = ki. Hence z = 4 ~ 2 are the only zeros of f(z) in the unit disc.

5413

(a) How many roots does this equation

z 4 + z + 5 = 0

have in the first quadrant? (b) How many of them have argument between f and $?


(a) Let R be sufficiently large such that when IzI = R,

Set f (z ) = z4 + z + 5

g ( z ) = z4 + 5. and

Choose a closed curve

lr I' = { z = z +iy;O 5 z 5 R,y = 0) U { z : IzI = R,O 5 argz 5 -} 2

U { z = z + i y : z = O , O < y L R}.

423

It is obvious that If(.) - g(z)I < Is(z.)l

holds when z E I?. By RouchC's theorem, the numbers of the zeros of f and g inside I' are equal. Since g has only one zero inside I?, f has also one zero inside I'. Noting that R can be arbitrarily large, we know that

z4 + z + 5 = 0

has one root in the first quadrant.

zero. Set (b) Let R be sufficiently large such that when la1 = R, 9 is approximately

f ( z ) = z 4 + z + 5

and

and A A r3 = { z : IzI = R, - < argz 5 -}. 4 - 2

It is easy to see that Imf(z) > 0 when

f(0) = 5, and

f ( ~ i ) E {w : o < argw < E } , f (a$*) E {w : ?r - E < argw <

where E > 0 is very small. We also know that

Ar,argf(z) = Ar,argz4 + Ar,arg (1 + F) ,

where Ar,mgf(z) denotes the change of argf(z) when z goes continuously from Re<' to Ri along I'3. It is obvious that

Ap,argz4 = A,

while

424

is very small. Let I' = I'l U r2 U r3 is taken once counterclockwise, it follows from the above discussion that

By the argument principle, the number of the roots of f (z ) = 0 inside equal to

is

Hence f(z) = z4 + z + 5 = 0

has exactly one root in the domain

x x { z : - < argz < -}.

4 2

5414

Prove that the equation sin z = z has infinitely many solutions in C. (Indiana)

Solution. Let

f(z) = sinz - z

and z = z + i y , then f (z ) can be written as

For any fixed natural number n, choose a positive number t >> logn and a closed contour I' = rl U I'2 U r3 U r4 in the counterclockwise sense, where

425

Then we consider the image of r under w = f (2):

f(rl) = {w = u + i~ : -2(n + 1 ) ~ 5 u 5 - 2 7 ~ ~ ) v = 0)

with the direction from the right to the left;

1 2 f(rz) = {w = u + i v : ti = -2(n+ ~ ) T , o 5 w 5 - (et - e-t ) - t )

with the direction upwards; f(I’3) lies in the annulus

starting from

w = -2(n + 1)n + i

in the counterclockwise sense;

with the direction downwards, Hence the winding number of f(r) around w = 0 is 1. By the argument

principle, f ( z ) = sinz - z has one zero inside the contour I?. Since n is arbitrarily chosen, we conclude that sinz = z has infinitely many solutions in c.

Remark. This problem can also be proved by Hadamard’s theorem. Assume that

f (z ) = sinz - z

has only finite zeros in (X, and denote all the zeros by z1, zz , . * , zn , multiple zeros being repeated. By Hadamard’s theorem, f (z ) can be written as

f(z) = eg(z) P k ) )

where n

P ( Z > = - zg) k=l

and g( z) is a polynomial.

426

It is obvious that f ( z ) is an entire function of order X = 1, where

T - 0 3

which implies that g ( z ) must be a polynomial of degree 1. Hence we have

sin z - z = ear+bp(z) .

Let z = z+iy and z be fixed. By letting y + +oo and y -+ -oo respectively, and comparing the increasing order on both sides, we obtain that Imu < 0 in the former case and that Imu > 0 in the latter case. This contradiction implies that sin z = z has infinite many solutions in C.

5415

(a) Let f be a non-constant analytic function in the annulus (1 < IzI < 2) and suppose that I f I = 5 on the boundary. Show that f has at least two zeros.

(b) I f f is meromorphic in the annulus, is the statement in part (a) still true?

Solution. (a) Let D = { z : 1 < It1 < 2) and d D = rl U r2, where rl = { z : IzI = 2)

is in the counterclockwise sense, and I'2 = { z : IzI = 1) is in the clockwise sense. Because f is non-constant analytic in D and I f 1 = 5 when z E dD, we know that both f(I'1) and f(rz) must be {w : lwl = 5) in the counterclockwise sense. Hence &Ap,argf(z) 2 1 and &Ar,argf(z) 2 1. In other words,

(Stanfod)

which shows by the argument principle that f has at least two zeros in D. (b) If f is meromorphic in D, the statement in (a) is not true. It might

occur that f(rl) and f(r2) are two subarcs of {w : IwI = 5), or both f(r1) and f(r2) are {w : IwI = 5) in the clockwise sense. In the latter case, f has no zero in D. The following is a counterexample. Let g(C) be a conformal map of

{ ( = < + i r l : $ + $ < l ,

427

onto {w : IwI > 5) with the normalization g(0) = M, g'(0) > 0. Then

is a non-constant meromorphic function in D with I f I = 5 when t E LID. But f has no zero in D.

5416

Let n be a positive integer, and let P be a polynomial of exact degree 2n:

P ( Z ) = a0 + alz + a2z2 + e . 0 + aZntzn,

where each aj €67, and aZn # 0. Suppose that there is no real number x such that P ( z ) = 0, and suppose that

Prove that P has exactly TI roots (counted with multiplicity) in the open upper half plane { z E 67 : Imz > 0).

(Indiana) Solution.

Let T > 0 be sufficiently large such that when I z I = T ,

( a 2 n ~ 2 n ( > (ao+alz+.. .+azn_1z2"-11.

Take a closed contour I? = I'l U I'2 in the counterclockwise sense, where

rl = { Z = TP : o 5 8 L .)

and r2 = { z = X + i y : -T 5 T,y = 01.

Then the number of zeros of P ( z ) inside I' is equal to

It is already known that

428

We also have

d log P(z) = G A r , 1 argP(z)

Note that 1

-Ap,arg (u2nz2n) = n 2a

and

we obtain that P has exactly n roots (counted with multiplicity) in the open upper half plane.

5417

Consider the function 1 1 1 1 1 z 2!z2 n! zn

f (z) = 1 + - + -- + .* .+ --.

(a) What does the integral

count? (b) What is the value of the integral for large n and fixed r? (c) What does this tell you about the zeros of f (z ) for large n?


(a) Let

1 1 1 1 c F(C) = f(-) = 1 + c + 3c2 + ,e3+ * . - + gcn.

From

429

we know that the negative of

represents the number of zeros of F(C) in (111 < :}, which is just the number of zeros of f (z) in (1.1 > T } .

(b) When n -+ 00, F(C) converges to ec uniformly in any compact subset of@. Let

min leCl = m, IC l=?

then m > 0. When n is sufficiently large,

lF(C) - ec I < m I 1ec I for ICI = :, which implies the numbers of zeros for F(C) and eC in { / ( I < :} are equal. Since ec has no zero in a', we obtain

for fixed T and large n. (c) From the above discussion, we conclude that for any fixed T > 0, when

n is sufficiently large, there is no zero of f ( z ) in { IzI > T } . In other words, all the n zeros of f ( z ) are in { IzI 5 T } .

5418

(a) Suppose that f ( z ) is analytic in the closed disk IzI 5 R, and that there is a unique, simple solution z1 of the equation f (z) = w in (1.1 < R}. Show that this solution is given by the formula

(b) Show that, if the integer n is sufficiently

W

large, the equation

has exactly one solution with ( z ( < 2.

430

(c) If z1 is the solution in (b), show that

1 1

n-cc 2 lim (zl - 1)- = -.


(a) Let f (z ) - w = ( z - zi)&(z),

where Q ( z ) is analytic and has no zero in { IzI < R}. Then

1 &’ (z ) f’(z) = [log(f(z) - w)]’ = [log(z - z ~ ) + log&(~)] ’ = - +- f - w z - z i Q ( z ) *

Hence

(b) Let fn(z) = z - 1 -

rE = and

= 2 - €1. For fixed large n, we choose E > 0 sufficiently small such that when z E re ,

z n E Ifn(z) - g(z) l= 151 = (1 - 5)” < 1 - E L Ig(z>I.

Hence fn(z) and g ( z ) have the same number of zeros in { IzI < 2 - E } , and the number is 1. Since E can be arbitrarily small, the equation z = 1 + ( 5 ) ” has exactly one solution (denoted by z p ’ ) in (1.1 < 2).

(c) f,(z) is a continuous real-valued function for 1 5 2 5 $. When n is sufficiently large, we have fn (1) < 0 and fn ( z ) > 0.

Hence we have z?’ E (1, $). It follows from

that

43 1

which implies

5419

Let 1 1 2 2

R = D(O,l)\{-, --}.

Find all analytic functions f : S2 + R with the following property: if y is any cycle in R which is not homologous to zero (mod R), then f * y is not homologous to zero (mod R).

Solution.

must be the removable singularities o f f . Let

(Indiana)

Since f is analytic in R and bounded by lf(z)I < 1, the points z = =kf

where E > 0 is small, and the directions of 71 and 7 2 are both in the counterclockwise sense. Since 71, 7 2 are not homologous to zero (mod R), f * yl and f *y2 are also not homologous to zero (mod R). As E tends to zero, f(rl) and f(y2) will tend to either 20 = 5 or 2u = -;, because otherwise, f * 71 or f * y2

will be homologous to zero (mod R). Hence we obtain

1

1 1 2

f(*-) = f2.

Now we claim that the case that f(i) = f(-;) will not happen. If, for example,

1 1 1 2 2

we assume that z = f is a zero of f(z) - f of order n and z = -f is a zero of f (z) - $ of order m, then

is homologous to zero (mod Q), while my1 - ny2 is not homologous to zero (mod R), which is a contradiction. Thus we obtain either

f(-1 = f(--) = 5,

f * (my1 - nr2)

1 1 1 1 f(-) = - f(--) = --

2 2 ' 2 2

432

or 1 1 2 2

f(-) 1 = --, 1 f(--) = -. 2 2

we consider the function 1 2 fk)- t . - -

1 - i f ( z ) 1 - t z F ( z ) =

which is analytic in D(0,l) and satisfies IF(z)I 5 1. It follows from F ( - i) = 1 that F ( z ) = 1, which implies that f (z) = z.

we consider the function f ( z ) + i z - 5 1

1 + if(.) - 1 - zz - -

1 G(z) =

which is also analytic in D(0,l) and satisfies IG(z)I 5 1. It follows from G ( - i ) = -1 that G(z) = -1, which implies that f(z) = -z. Thus we conclude that the functions which satisfy the requirements of the problem are f (z) = z and f(z) = -2.

433

SECTION 5 SERIES AND NORMAL FAMILIES

5501

00 Let

n=O

have a radius of convergence T and let the function f ( z ) to which it converges have exactly one singular point zo, on It1 = T , which is a simple pole. Prove that

lim an/an+l = zo. n-+w

Solution. Assume that the residue of f ( z ) at zo is A , and define

(Indiana)

A z - zo

F ( z ) = f ( z ) - -.

Then F ( z ) is analytic on { z : 1zJ 5 T } . In other words, the Taylor expansion of F ( z ) a t z = 0 has a radius of convergence larger than T . Hence the power series

A 00

F ( z ) = C a n t n - - z - zo

n=O

is convergent a t z = zo, which implies

It follows that

434

and

and we obtain a n lim - = zo.

n-+co % + I

5502

(1) Show that the series

n> 1

is convergent for 1 # a EC with la/ = 1. (2) Show that this series converges to log(1 - a ) for such a.


(1) Let a = eit , t E ( 0 , 2 ~ ) , then

-c,=-c an cos nt + i sin nt n

n > l n > l

For t E ( 0 , 2 ~ ) we have

and

Because both C for 1 # a EC with la1 = 1.

tends to zero monotonically, by Dirichlet’s criterion we know that converge, which shows that - C $ is convergent and C

n > l n > l n > l

(2) Let zn n f ( z ) = - c - (1.1 < 1).

n > l

Differentiating term by term, we have

-1 1 - Z

n > 1

435

Integrating both sides on the above identity, we obtain f(z) = log(1- z ) , for

Let Q = eit, z = Teat where 0 < T < 1, 0 < t < 2a. It follows from Abel's (a1 < 1.

limit theorem that

-C- an = -C,= eint l i m - C - (Teit)n r-+1- n

n> 1 n > l n

n> 1

= lim log(1- Te i t ) = log(1- e i t ) = log(1 - a). r+ l -

5503

Consider a power series O01 I C -zn. .

n n= 1

Show that the series converges to a holomorphic function on the open unit disk centered at origin. Prove that the boundary of the disk is the natural boundary of the function.

Solution.

of

( Columbia)

First of all, we prove the following proposition: If the radius of convergence

n =O

is equal to 1 and an 2 0 for all n, then a = 1 is a singular point of f(z). Assume the proposition is false, i.e., z = 1 is a regular point of f, then for fixed 2 E (0 , l ) there exists a small real number 6 > 0 such that the power series expansion of f at point 3: is convergent at z = 1 + 6. Suppose the series is

436

Thus 00 w w n(n - 1) . . a ( . - Ic + 1)

x b k ( z - z ) k = &I an(z - z )kzn-k n,.

k=O k=On=k

is convergent at z = 1 + 6 . Noting that when z = 1 + 6 the right side in the above identity is a convergent double series with positive terms, and hence the order of summation can be changed, we assert that when z = 1 + 6 ,

." . n=O k=O W

n=O

which contradicts the statement that the radius of convergence of

n=O

is equal to 1. Now we return to the power series

It follows from

that the radius of convergence of

c " 1 -zn!

n

is equal to 1. By the above proposition, z = 1 is a singular point of F ( z ) . For any natural numbers p and q,

n = l

437

Since z = 1 is a singular point of

" 1 I c -zn., n

n = p

it is also a singular point of F ( z e 9 " ' ) . In other words, z = e-?4,i is a singular point of F ( z ) . Since the set { e - p r i : p , q = 1,2,...} is dense on {It1 = l}, we conclude that the unit circle {IzI = 1) is the natural boundary of F ( z ) .

Remark. By the above discussion, the boundary of the unit disk is also the

natural boundary of the function C $zn! although the series is absolutely

and uniformly convergent on the closure of the unit disk.

aq

00

n=l

5504

Suppose f is analytic in U = (1.1 < 1) with f(0) = 0 and If(z)I < 1 for all z E U . If the sequence { f n } is defined by composition

and f n ( z > 4 g(z)

for all z E U , prove that either g ( z ) = 0 or g ( z ) = z. (Indiana-Purdue)

Solution. By Schwarz's lemma, it follows from f(0) = 0 and If(z)l < 1 that If(.)[ 5

IzI for all z E U, and if If(z)I = 1.1 for some z # 0, then f(z) = einz where a is a real number.

In the case when f (z ) = eiaz. f n ( z ) = einaz. Since f n (z ) is convergent, we obtain a = 0, which implies that f (z ) = z and g(z) = t.

In other cases, we have

for all z E U . Let 0 < r < 1. Then

438

For all z E { IzI 5 T } , we have

. . .

..

Hence fn(z) converges to zero uniformly in-{lzl 5 T } . Since 0 < T < 1 is arbitrarily chosen, we obtain g ( z ) = 0 for all z E U .

5505

Let {fn}r=l be a sequence of analytic functions in a domain D which converges uniformaly on compact subsets of D t o a function f on D.

(a) Prove that if fn(z) # 0 for all n 2 1 and z E D, then either f is identically zero in D or f(z) # 0 for all z E D.

(b) If each fn is one-to-one on D , show that f is either constant or one-to- one on D .

(UC, Irvine) Solution.

(a) First of all, we know from Weierstrass’ theorem that f is analytic on D . Suppose f is pot identically zero, but has a zero point zo E D. Since the zeros of a non-zero analytic function are isolated, there exists T > 0, such that f(z) # 0 when

z E {Z : 0 0. As {fn} converges to f(z) uniformly on compact subsets of D , we know that for sufficiently large n,

Ifn(z> - f(z>I < m 5 l f ( z> I

holds on {z : Iz - z0I = T } . It follows from RouchC’s theorem that fn and f have the same number of zeros in { z : Iz - z0I < T } . Since zo is a zero of f , fn must have a zero in { z : Iz - z0I < T } . which is a contradiction to the assumption that fn(z) # 0 for all z E D.

439

(b) Suppose f is not a constant, and is not one-to-one on D. Then there exist zl,z2 E D (z1 # zz), such that f(z1) = f(z2) (denote it by u). Choose T > 0 sufficiently small, such that

and f (z ) - a # 0 in {z : 0 < (z - z l ( 5 T } U {z : 0 < Iz - 221 5 T } . Let rn be the minimum value of If(.) - a1 on { z : I Z - zll = T or Iz - z2) = T } . Then m > 0. With the same reason as in (a), when n is sufficiently large,

I(fn(z) - a) - (f(z) - .)I = Ifn(z) - f(z)I < m L If(.) - al

holds on { z : Iz - zll = T or (z - z2 ( = T } . It follows from RouchC’s theorem that f n (z) - a and f(z) - a have the same number of zeros in {z : )z - z11 < T }

and { z : (z - 221 < T } respectively. In other words, there exists

2: E { z : Iz - 211 < T }

and z: E { z : (z - z2( < T } ,

such that fn(Z{) - a = 0 and fn(zi) - u = 0, which implies fn(z{) = fn(zi) (2: # 24). This is a contradiction to the assumption that fn is one-to-one on D.

5506

Let D c a! be a bounded domain, and let { f n } be a sequence of analytic automorphisms of D such that

lim fn(a) = b E 8D n - + m

for some point u E D. Prove that

lim f n ( z ) = b n+m

for every z E D. (Indiana)

440

Solution. Take a0 E D , ao # a. If {fn(ao)} does not converge to b, there exists a

subsequence of {fn(ao)} converging to bo # b. Without loss of generality, we assume

lim fn(u) = b 6 dD,

lim fn(ao) = bo # b. n-Pm

n+ca

Since { f n ( z ) } is a normal family, there is a subsequence { f n k ( z ) ) converging uniformly on compact subsets of D to f (z) . Because f(u) # f ( a o ) , f (z) is a non-constant analytic function of D .

Let T be sufficiently small such that f ( z ) - b has no zero in { z : 0 < Iz -u[ 5 T } c D, then m = min{lf(z) - bl : )z - a1 = T } > 0. Since { f n k } converges uniformly to f on { z : ( z - a1 = T } , when Ic is sufficiently large,

on { z : Iz - a1 = T } . By Rouche’s theorem, fnk(z) - b has zero(s) in { z : Iz - a1 < T } , which is a contradiction to the fact that fnk does not assume the value b E d D in D because f n k is an automorphism of D .

5507

Which of the following families are normal, and which is compact? Justify

(a) 3 = {f : f is analytic in D , f(0) = 0, diam f ( D ) 5 2) (b) B = { g : g is analytic in D,g(O) = l ,Re{g} > 0, diam g(D) 2 1).

Here the diameter of a set S is diam S = sup{lz - CI : z , C E S}.

S o h tion. (a) For any f E F, it follows from f(0) = 0 and diam f ( D ) 5 2 that

lf(z)I 5 2, which shows that 7 is normal. Let {fn} be a sequence of functions in F. Then there exists a subsequence

{ f n k } converging uniformly in compact subsets of D to f(z), which obviously satisfies the conditions that f(z) is analytic in D and f(0) = 0. For any two fixed points z,C E D , we have

your answers.

(Indiana)

441

because diam fn , (D) 5 2. We choose a compact subset K c D such that z,C E K . It follows from the uniform convergence of {f,,} on K that

If(.) - f(0l 52.

Since z , c E D can be arbitrarily chosen, we obtain diam f ( D ) 5 2, hence f(z) E 7, which shows that F is also compact.

(b) Let { g , } be any sequence of functions in Q. Then for Gn(z) = e-gn('),

we have IGn(z)l < 1. Hence there exists a subsequence {Gnk} converging uniformly in compact subsets of D to a function G(z) which is either a constant or a non-constant analytic function in D . If G(z) is a constant, then the constant is e-l because

~ ( 0 ) = lim e-gn(') = e-1;

if G(z) is non-constant analytic, since G,(z) # 0 for all z E D , by Hurwitz's theorem, we have G(z) # 0 for all z E D . Hence we can define an analytic function g ( z ) = -log G(z), where the single-valued branch is chosen by g(0) = - log G(0) = 1, and we conclude that

n-vw

gn,( . ) = -logGn,(z)

converges uniformly in compact subsets of D to g ( z ) , which shows that family Q is normal. But family Q is not compact. First we can choose a sequence of functions g n ( z ) in Q as follows: g n ( z ) is a conformal mapping of D onto

1 1 4 n R, = {w : Iw - 11 < -} U {w : Iw - 31 < 1) U {w : IImwl < -, 1 < Rew < 3)

satisfying gn(0) = 1, gL(0) > 0. By the Riemann mapping theorem, such a mapping gn exists and is unique, and it is obvious that gn satisfies all the conditions required by the family Q. Because the domain sequence {a,} converges to R = {w : ( w - 11 < a} which is called the kernel of {R,} with respect to w = 1, by Caratheodory's theorem, { g n ( z ) } converges uniformly in compact subsets of D to g(z ) which is a conformal mapping of D onto 52. Since diam g(D) = i, g ( z ) does not belong to the family 0, which shows that is not compact.

5508

Suppose that 1 5 p < 00 and c 2 0 is a real number. Let 7 be the set of all analytic functions f on { IzI < 1) such that

442

sup /'^ If(reioe)lPde 5 c. O<r<l o

Show that 3 is a normal family. (Illinois)

Solution. It suffices to prove that the functions in 3 are uniformly bounded on every

compact set of { z I < 1). We prove the assertion by contradiction. If it is not the case, then there exist zn E D, f n E F such that zn --+ zo E D and

Let 1 - lzol = 3r. Then when n is sufficiently large, Izn - 201 < r. By f n ( Z n ) + 00.

Cauchy integral formula,

Hence

where

Then

1 1 - + - = l . P q

As n + co, the left side of the above inequality tends to infinity, while the right side of the inequality is a constant. The contradiction implies that 3 is a normal family.

443

5500

(a) Let f be holomorphic for 121 < R and satisfy f(0) = 0, f'(0) # 0, f (z) # 0 for 0 < IzI < r 5 R. Let C be the circle IzI = p where p < T . Show that

1 t f ' ( t ) d t g(w) = 2?ri J, f ( t ) - w

define a holomorphic function of w for

1wI < m = m p If(pe")l,

and that z = g(w) is the unique solution of

that tends to zero with w.

series expansion of the root of the equation (b) Find the Taylor's expansion of g(w), and apply this to find the explicit

z3 + 32 - w = 0

that tends to zero with w.

Solution. (a) It follows from

IwI < m = rnp lf(pe")l

that when t E C,

Hence

=5 (&J ,p t f ' ( t ) d t ) wn, g(w) = GJ, f ( t ) - w n=O

1 t f ' ( t ) d t

which implies that g(w) is holomorphic in {w : IwI < m}.

444

Let r be the image of C under f where C is the circle { z : IzI = p } taken once counterclockwise. Because

lwl < m = min I f (pe”)l ,

@ , O ) = n(r, w),

9

the winding number

which shows that f(z) and f ( z ) - w have the same number of zeros in { z : IzI < p } . Since z = 0 is the only simple zero of f in { z : IzI < p } , we know that f (z) = w has a unique solution in { z : 1x1 < p} . Denote the unique solution by z l , then

f(t) - w = (t - zi)Q(t)

where Q(t) is analytic and has no zero in {t : It1 = pog(f(t) - w)]’ = [log(t - z1) + log Q(t)]’ = - + - f(t) - w t - zi Q(t) *

f’(t)

Hence

which shows that g(w) is just the unique solution of f (z) = w. As the constant term in the Taylor expansion of g(w) is

which is obviously zero, we assert that the unique solution g(w) tends to zero together with w .

(b) Let f (z) = z3 + 32,

then

where

445

After some computation, we obtain a 2 k = 0 and

a2k-1 = z l 1 3 2 k - l t 2 k - 1 [ 2 '-2k k - 2 (:)k-2 + '-2k k - 1 (5)k-'] dt

- - 33k-2 ( 3ck-2 -2k t c",:) .

5510

Find an explicit formula for a meromorphic function f whose only singularities are simple poles at -1, -2, - 3 , - - - with residue n at z = -n. Prove in detail that your function has all the required properties.


By Mittag-Leffler's theorem, we construct

For any natural number N , when IzI 5 N , n 2 2N,

2N2

Hence

converges uniformly in { IzI 5 N } to a function which is analytic in { IzI < N } . In addition.

z2 2N-1

n = l

is a meromorphic function whose only singularities in { IzI < N } are simple poles at z = -1, -2, * - . , - N + 1 with residue n at z = -n. So f ( z ) is analytic in (1.1 < N}\{- l , -2 , . . . , -N+1} , andz = -1,-2,...,-N+lareitssimple poles with residue n at z = -n.

Because N can be chosen arbitrarily large, it is obvious that f ( z ) has all the required properties of the problem.

446

5511

(a) Does there exist a sequence of polynomials {P,} such that Pn(z) --i 5 uniformly on the annulus 1 < IzI < 2? If Yes, give an explicit formula for the Pn ; if No, explain why not.

(b) Does there exist an entire function g whose zero-set is {+( 1 + i) : n = 0,1,2,3, . . -)? If Yes, give an explicit formula for g; if No, explain why not.


(a) No. If there exists asequence of polynomials {P,} such that Pn(z) -+ 5 uniformly on (1 < 1.1 < 2} , then for any E E (0, a), there exists N > 0 such that when n > N , IPn(z) - $ 1 < E holds for all z E (1 < Izl < 2). Multiply both sides by )zI2, we have

Because z2Pn(z ) - 1 is an analytic function in (1.1 < 2}, it follows from the maximum modulus principle that

holds for all z E { Izl < 2 ) . The contradiction follows by taking z = 0 in the inequality.

(b) Yes. The function g can be chosen as

where an = fi(1 + i). For any R > 0, let (z( 5 R and choose N > R2. Then when n 2 N ,


447

00

Because C 5 converges, we know that n=N

is analytic in {z : IzI < R}, which implies

m

is analytic in { z : IzI < R } . Hence g(z) is analytic in {z : IzI < R}, and its zeros in {z : 1z1 < R} are 0,al,a2,...,ak (q - 1 5 IC < $). Since R can be arbitrarily large, we see that g(z) is an entire function with the required zereset.

5512

State whether the following statement is True or False, and prove your

For each positive integer n there exists an entire function fn such that assertion.

(Indiana) Solution.

False. We prove the assertion by contradiction. If for each positive number n there exists an entire function fn such that

then for 15 IzI 5 2, we have

-1 5 Refn(z) 5 1+log2.

Define Fn(z) = efn('). Then Fn(z) are entire functions with no zeros, and

1 - 5 ~ ~ ~ ( z > l = eRefm(z) < - 2e e

448

for 15 IzI 5 2. By the maximum modulus principle, IF,(z)I 5 2e for IzI 5 2. Hence {F,(z)} is a normal family in { z : ( z ( < 2), and there exists a subsequence {Fnk(z)} converging locally uniformly to an analytic function F ( z ) in { z : IzI < 2). Since IFn(z)l 2 for 15 IzI 5 2, F ( z ) cannot be identically zero, and by Hurwitz's theorem F ( z ) has no zero in { z : 1x1 < 2). But we have for 15 IzI < 2,

which implies that F ( z ) = cyz with contradiction to the fact that F ( z ) has no zero in { z : IzI < 2).

= 1 in { z : It1 < 2). This is a

5513

Let G = D\(-1,0], where D = { z : IzI < 1). (a) Give a single-valued definition for zi in G. (b) Why should there exist a sequence of polynomials P, such that

lim P,(z) = z* n-rm

for all t in G?

IPn(z)l 5 M for all z E G and all n? Justify your answer.

Solution. (a) zi is defined by eil0gz. In domain G, single-valued branch of logz can

be chosen. For example, a single-valued branch of zi in G can be defined by

(c) Can the polynomials be chosen so that there exists a constant M with

(Indiana)

argzlo<2<l = 0. (b) Choose

where n > 2. Then K, C Kn+lr and

lim K, = G. n-rco

Because the complement of K, is connected and contains z = 00, we know by Runge's theorem that there exists a sequence of polynomials which converges

449

uniformly on K , to z i . In other words, we can find a polynomial Pn(z) such that

for all z E K,. Hence {Pn(Z); n = 1,2, . . - } converges to zi uniformly on compact subsets of G.

(c) No. If there exists M with IPn(z)l 5 M for all z E G and all n, then because Pn(z) are continuous on D, IPn(z)l 5 M for all z E D and all n. It follows that {Pn(z)} is a normal family in D, and there exists a subsequence P,, ( z ) which converges uniformly on compact subsets of D to an analytic function f (z ) in D. Since Pn(z) converges to zi in G, hence zi = f (z) for z E G, which implies that z' can be extended to a single-valued analytic function in D. It is obvious impossible, so the contradiction is obtained.

5514

(a) Prove that

(b) Use this to show that

l o o 2z a c o t a z = - + c - Z 2 2 - n2 '

n = l

Justify your steps.

use of (b), with enough terms to find the values of C 5 and C (c) Develop a cot az in a Laurent series about the origin directly and by

00 00 1

n = l n = l (Hamad)

Solution. (a) Let

a2 f (z) = G.

1 The singular part of f at z = n (n = 0, fl, f2, - * .) is - (2-n)2. Now we consider the series

450

For any natural number N and la1 5 N,

00

holds for n 2 2N and n 5 -2N. It follows from the convergence of C 5 n=2N

n=-2N M ~~

and C 5 that C & is analytic in -cQ n=-m

Because N can be arbitrarily large, we obtain the result that

1 00

n=-cc

is a meromorphic function which has the same singulaxities as f(z). Let

Then g ( z ) is an entire function. As f (z) and

are both periodic functions with period equal to 1, we restrict z in the strip

{ z : 0 < Rez 5 1).

It is obvious that lim f (z ) = 0.

Imt+f00

As the convergence of

is uniform for lIm4 2 1,

the limit of the series for Imy -+ fco can be obtained by taking the limit in each term and the limit is also zero. Hence g(z ) is a bounded entire function,

45 1

which implies that g ( z ) is a constant. It is obvious that the constant must be zero. Thus we obtain the identity

(b) Let F ( z ) = 7rctg7rz.

The singular part of F at z = n (n = 0, fl , &2, - - - ) is A. Now we consider the series

1 " 22

n = l

With similar discussion to that in (a), we know that

is a meromorphic function which has the same singularities as F ( z ) . Let

22

22 - n2 + G(z) . 1 "

F ( z ) = ; + c - n = l

Then G ( z ) is an entire function. Differentiating both sides of the above identity, we obtain

- G'(z) T 2 1 " 1 1

n = l

Comparing this identity with (l), we have G'(z) = 0 which implies that G = c ( c is a constant). For

+ c, z2 - n2 n = l

it follows from the fact that F ( z ) and

1 " 22 - + E n n = l

are both odd functions that c = 0. Hence we obtain

1 " 22

n = l

7rcot?fz= -+c- z2 - n2'

452

(c) The Laurent expansion of 7rcot 7rz about the origin is

1 7r2 7r4

z 3 45 jl. cot r z x - - -z - -23 - . . . .

It follows from (2) and (3) that around the origin,

(3)

Take z = 0, we obtain

After differentiating (4) on both sides, we can also obtain

5515

Let z1, - - , Zn be distinct complex numbers. Let f and g be polynomials, f of degree 5 n - 2 and

g ( z ) = ( z - tl) 9 * - ( z - zn).

(a) Show that

(b) Show that there exists a polynomial of degree 5 n - 2 with f (zj) = uj

if and only if n

(c) Given a sequence of complex numbers z1,z2,. . - such that lzn 1 + 00,

does there exist an entire function f with f(zj) = aj? Can you write this function down?

(Hamad)

453

Solution. (a) Take R sufficiently large such that

Z l , Z 2 , " ' , Z , E (1.1 < RS.

g ( z ) = ( z - q ) ( z - z 2 ) * . . ( z - Z n )

Because

is of degree n, while f ( z ) is of degree 5 n - 2,

-dz = lim -dz = 0.

we obtain

? # = O . j = 1

(b) If f ( z ) is a polynomial of degree 5 n - 2 with f(zj) = uj ( j = 1,2,. - . , n), then by (a), we have

If u l , u 2 , . - , an are n complex numbers such that n

we construct the function f ( z ) by

For each j ,

-- g ( z ) - (2 - z1) . . . ( z - z j - l ) ( z - Z j + l ) . . . ( z - z,) z - zj

is a polynomial of degree n - 1, and the coefficient of zn-' is 1. Since

454

the coefficient of zn-l of f(z) is zero. In other words, f(z) is a polynomial of degree 5 n - 2.

Because aj d z ) -

- a j , lim -. ~

z + r j g ’ ( z j ) ( z - Zj)

while for k # j ,

f ( z ) satisfies the condition f(zj) = aj ( j = 1,2, (c) For the given sequence z1, z2, - a , such that 11, I -+ 00, by the Weierstrass

theorem about the canonical product of entire functions, we can construct an entire function g ( z ) with simple zeros z l , z2,. - -. Then we define

- , n).

where 7, is chosen such that when IzI 5 k$,

Because Iz,I -+ 00, for any R > 0, there exists N > 0 such that Iz,I > 2R when n 2 N . Hence

1 Iun(z>l 5 2

00

holds for all IzI 5 R when n 2 N . In other words, un(a) converges uni-

formly for all IzI 5 R, so that f (z) is analytic in (1.1 < R}. Since R can be arbitrarily large, f ( z ) is an entire function.

n = l


while for k # n, u k ( z n ) = 0,

which implies that f ( z ) is an entire function satisfying the required condition.

Part VI

Partial Differential Equations

457

SECTION 1 GENERAL THEORY

6101

a) Let A = (a,j)&=l be a real matrix. Show that

W n ( 2 , Ax)dx = trace A.

L<l n(n + 2)

Here (., -) is the dot product of vectors in R" and wn is the area of the unit sphere in En.

b) Show that u E C:(R") implies

(Iowa) Solution.

a) It is not difficult to verify that

aijxjxjdx = 0, Vi # j lx l < l

and

Therefore,

1 n

n(n + 2)

= - trace A

trace A. - W n -

458

b) First let u E Cr(IRn). By applying Green's formula, we get immediately

As Cr(lRn) is dense in Hi(IRn) , the conclusion is true for u E Ci(IRn).

6102

Let T be a distribution on IR and suppose that T' = 0 on IR. Show that T = const; i.e., show that there is a number a such that

T(q5) = / aq5dx for all q5 E Cr(lR) . R

(Indiana) Solution.

T' = 0 if and only if

T(q5') = 0, Vq5 E C r ( B ) .

It is easy to verify that a function q5 in Cr(nZ) is the derivative of a function in C r (B) if and only if

q5dx = 0. IR

Take a function p E Cr( lR) such that

pdx = 1. J, Then it is easy to verify that for any q5 E Cr(IR)

satisfies the condition ?

J, g d x = 0.

459

Hence T($) = 0 and

6103

Let u E Hs(En) with s > n/2. Show that lim u(z) = 0. 14+w

( cin cinn at2) Solution.

We show first that 6 E L 1 ( B n ) if u E H s ( E n ) with s > n/2. In fact, u E H s ( R n ) if and only if (1 + lt12)'/2G E L 2 ( R n ) . And it is clear that (1 + 1 [ 1 2 ) - a / 2 E L2(Rn) if s > n/2. Therefore, we have 6 E L 1 ( R n ) .

Then we get immediately

6104

Let u be defined for f E D((0,l)) by

Determine whether u is a distribution on (0, l), and support your answer.

Solution.

be a sequence in D((0,l)) such that

(Indiana)

It is clear that u, defined above, is a linear functional in D((0,l)). Let {fk}

fk + 0 in D((0 , l)), as Ic -+ 00.

Then we have a compact interval [a, b] c ( 0 , l ) such that

SUPPfk c [a, bl, vk:

460

and jp) -, o uniformly on [u, b] as -+ oo,

for any nonnegative integer n. Let N be a positive integer such that

1 - < a. N

Then we have

Therefore, u is continuous in D ( ( 0 , l)), and is a distribution on (0, l ) .

6105

Let B = {(x, y) I x2 + Y2 < 1).

For which p 2 1 does the function

5

belong to WIJ’(B)?

Solution. Let

XY uy(x7 y, - (.2 + y2)3/2 *

It is clear that us,uy E L1(B).

in the sense of distributions, that is

(Iowa)

J, u#,dxdy = - u,#dxdy J,

461

and r r

u+ydxdy = - JB uy#dxdy JB

for all radius E < 1. Let RE = B\BE and S, = aB,. Then we have

E Cr(B). By B, we denote the ball centred at the origin with the

u4,dzdy = - lc ux4dxdy + 1. u ~ c o s ( ? ~ ' , x)ds J,. for any + E C r ( B ) . Letting E -, 0 in the above inequality, we get &u = u,. Similarly L u = 2ly. ' BY

It is not difficult to verify that

for 1 5 p < 2. Therefore, u E W1ip(B) for 1 5 p < 2. And we can verify that

which means that u @ W1yP(B) for p = 2.

6106

Let C be a smooth hypersurface dividing R" into two disjoint open regions Denote by v = v(x) the unit normal to C pointing into R2. Q , and R2.

Suppose v : R" -+ R" solves the equation

divv(x) = b(x) for x E R" - C ( *I in the classical (C') sense where b is a continuous function. Assume furthermore that v lies in both Co(?&) and C0((n2) (but not necessarily Co((IRn)!!) with

v+(xo) lim v(x), vu-(xo) lim v(x) x-xo x-DO

+En1 X E n *

for each xo E C . Derive a necessary and sufficient condition in terms of v+, v- and v for ZI to be a distribution solution of (*) on all of R".

Solution. (Indiana)

v is a distribution solution of (*) in En if and only if

462

v . V#dx = IRn b#dx, V# E CF(lRn). - J,.

2 , . V+dx = - J,, 2 , . V#dx - J,, 2 ) . V+dx -Ln lRn

By Green’s formula, we have

= - +(v+ - v-) uds + divv . #dx.

Hence the equation (1) is equivalent to

#(v+ - V-) . vds = 0, V# E C;(IRn).

Therefore v is a distribution solution to (*) on all of B* if and only if

(w+ - v-) - Y = o on C.

6107

Recall the definition of the curl operator in two dimensions acting on a vector field U(z,y) = (u(z,y),v(x,y)):

v x u u ~ 2 L y - v u , .

(a) Give a definition of what it means for a (not necessarily continuous)

(b) Suppose U takes the form vector field U to have curl zero in the sense of distributions.

where 521 and a2 are open sets such that a1 U n2 = B2, I’ z d a l n a522 is a smooth curve, and U1 and U2 are (different) constant vectors. If U has curl zero in lR2 in the sense of distributions, describe as completely as possible the curve r.

Solution. (Indiana)

(a) Let u,v E Lloc(B2). Then

curl U = 0

463

in the sense of distributions is defined as

(b) Let 171 = (u1, WI), U2 = (uz, vz), where u1, w1, u2 and 712 are constants. Then curl U = 0 if and only if

Integrating by parts, we have

J,(ul - %)4dz + (v1 - v2)4dy = 0, v4 E C,-(lR”.

Therefore, I’ is a straight line defined by

(u1 - u2)dz + (v1 - v2)dy = 0.

6108

Let p be a polynomial in n variables, and assume that the set {z : p ( i z ) = 0) is bounded. Prove that there exists a tempered distribution E on lRn such that p ( D ) E - 6 E S. Here S is the Schwartz class of rapidly decreasing functions, and 6 is the Dirac distribution defined by 6 - 4 = 4(0).

Solution.

such that

(Indiana)

As the set {< : p( i<) = 0) is bounded, there exists a positive constant R

{< : P(i<) = 0) c {< : 14 < R). Construct a function XR E Cp(lRn) such that

XR(<) = 1, v< E {< : < R}.

Set

It is clear that E E S’(Rn) n Cm(Rn), where S’(W) denotes the space of tempered distributions on B”, and that

464

It is easy to see that p ( i € ) Z ( € ) - 1 E C,-(nZ").

p ( D ) E - 6 E s, Therefore, we get

where E = F - ' ( Z ) E S'(nZn).

6109

Let P ( D ) be an mth order linear constant coefficient partial differential operator on En; i.e.,

P ( D ) = C,D"

where a denotes a multi-index Q = (a1, ~ 2 , . - , a,) and the C, are constants. Suppose

b l l m

~ ( i < ) E C ~,(i<>" # o for all < E - (0). I 4 < m

If u E S' satisfies P(D)u = 0, prove that u must be a polynomial. (Here S' denotes the space of tempered distributions on an.)

Solution. (Indiana)

Let G(E) be the Fourier transform of u. Transforming the equation, we get

P ( g ) G ( [ ) = 0.

From the transformed equation, we can see that

suppG = { 0).

Therefore, G ( t ) must be finite linear combinations of the derivatives of the Dirac delta function S(<) at { 0 } , i.e.,

.^(€I = c a,S(")(<), l 4 l N

where a = (a1, a2, + - . , a,) is a multi-index and a, are constants. Hence

is a polynomial.

465

6110

in D'. (b) Show how to define as a distribution (i.e., define the Cauchy Principle

Value and show that it is in D').

terms of your answer to (b). (c) Calculate rigorously the derivative of log 1x1 and express the result in

(d) Calculate the Fourier transform of the distribution 5. (Indiana)

Solution. (a) It is well-known that

e-xaf4tdx = 1 Vt > 0.

For any 4 E D, we have

Then it is not difficult to verify that

Hence e-x2/4t - 6 lim - - ( 1.

t+O+ & (b) We define the distribution 5 by its Cauchy Principle Value &: as

follows:

The linearity of the functional Vp: is obvious. Further, suppose that q5n -+ 0 in D. Let supp4, c ( -a ,a ) Vn. Then we have

466

Therefore, Vpi is a distribution in D'. ( c ) For any 4 E D, we have

Hence -&log 121 = V,:.

For any 4 E S, we have (d) By F ( f ) we denote the Fourier transform of a tempered distribution f

Here, we have applied the integral equality

'J- -sin(zt)dt = 1 for z > 0. 'IF -00t

Therefore, we obtain

6111

a) Let Lu(2) = c a,DcIu(c) 2 €mn

bI<m

467

where the abs are constants, and a is a multi-index. What is a fundamental solution of L? When is it unique?

b) If I+ denotes the Heaviside step function

Show that 1 2

is a fundamental solution of the wave operator W(u) = utt - u,, on E2.

F ( z , t ) = -I+(t)l+(t - 1.1)

c) Express the solution of the Cauchy problem

utt - u,, = #J(z , t ) , u(z, 0) = f(z), U t ( 2 , O ) = g(z),

-00 < 2 < 00, t > 0 -00 < 2 < 00

(f, g, #J sufficiently nice) in terms of the above fundamental solution. Simplify your result to obtain D’Alembert’s solution of this problem.

Solution.

ferential operator L if

(Indiana)

a) A distribution E E D’(nZn) is called a fundamental solution of the dif-

L E ( 2 ) = S(z),

where S(z) is the Dirac function. Let L* be the formal adjoint of L defined by

L * V ( C ) = c ( - l ) ~ a b a ( a a V ( z ) ) . l a l lm

If L*S is dense in S, where S is the space of rapidly decreasing functions, then the operator L has at most one fundamental solution in S‘.

b) We need only to show that

In fact

468

The solution of the Cauchy problem

is given by

Therefore, the solution of the original Cauchy problem can be obtained as

469

This is just the D’Alembert’s formula when q5 2 0.

6112

Let u be the solution of the initial problem

where summation is understood in the pde, and 4 is a C” function on IR3 with compact support. Assume that there is a constant C such that

(Indiana) Solution.

By applying Fourier transform to the problem, we have

ztt + i t 1 4 6 = 0,

qt, 0) = 0, q t , 0) = w, where G and 3 denote the Fourier transforms of u and q5 with respect to z, respectively.

By solving the above transformed problem, it yields that

% t ) =

Therefore, we have

Set 77 = tit in the integral in the right side of the above equality. It is reduced to

470

This implies that

In fact, if I&(O)l = 2a # 0, then there exists a positive constant r such that

Iml 2 a, K E E3, 14 < r.

And when t > 5, we have

This contradicts the condition satisfied by the solution u, given in the problem.

6113

Given E S (rapidly decreasing functions) over B1, consider the solution u of the SchrGdinger Equation

Show that lu(z,t)I2dz = I&0124-

Here 3 denotes the Fourier Transform of 4.

Solution.

heat equation applies to the Schr6dinger equation. So we have

(Indiana)

It can be verified that the Poisson's formula for the Cauchy problem of the

And then

47 1

Set x = 2&, we get

Here the Fourier transform of 4 is defined by

472

SECTION 2 ELLIPTIC EQUATIONS

6201

Let u $ 0 satisfy u E C2(Rn), Au = 0 on R". Show that

does not exist.

Solution.

value property for the harmonic function u, we have

(Indiana)

There exists a point xo E En such that u ( x o ) # 0. Applying the mean-

where w , is the surface area of the unit sphere in En. Schwarz's inequality gives

that is JT u 2 ( x ) d S z >_ w,pn-1U2(x0) . z - z * ( = p

Then we have

u 2 ( x ) d x

wnu2(xo)J lo pn-'dp

w n 2 0 n = -u ( x )T , n

473

for any T > 0. The conclusion follows as T + 00.

6202

Let u E Co(R) be weakly harmonic in an open set R, i.e., the relation

uAddx = 0

holds for all 4 E C,(R). Show that u is then harmonic in R. (Iowa)

Solution. Set

where C is a constant such that

For x E R, set 2 - Y

PE(Y) = E-P (7) Then it is easy to see that

P4Y) E C,-(f% provided E < dist(z, an). Therefore, we have

where

This implies that for any compact set K c R, u, is harmonic in K if E < dist(K, an).

It is well-known that

u, + u uniformly in K , as E -+ 0.

Therefore u is harmonic in R.

474

6203

Let f (2, y) be a locally bounded function, harmonic in z E lR2 and continuous in y E BIZ. Show that f is continuous in (2, y), i.e., as a function on B4.

Solution.

such that

(Indiana)

Let (zo, yo) E B4 and R > 0 be given. Then there exists a constant M > 0

lf(z,y)I 5 M, V(z,y) E lR4, 12 - z0l2 + Iy - y0l2 < 2R2.

Applying the mean-value theorem to the harmonic function z, we have

By Green's formula, we get

For z, we have the same estimation. These estimations imply that f ( z , y ) is continuous at z = z0 uniformly

with respect to y near yo. Then the conclusion that f (z ,y) is continuous at ( zo, yo) E B4 follows immediately.

6204

Suppose u is a function defined on nZn such that (i) u is bounded and continuous on the half-space xC 2 0 of En, (ii) u = 0 on z, = 0, (iii) u is harmonic in z, > 0.

Prove that u E 0 on z, 2 0. (Indiana)

475

Solution. Redefine the function u in the half-space x, < 0 as an odd function of 2,:

u(x‘,2n) = -u(x’,-x~), VX, < O

where x’ = (xl,e--,xn-l). By using the uniqueness theorem of the Dirichlet problem and the Poisson’s formula of the Dirichlet problem in a ball, we can verify that the redefined function u is harmonic in any ball centred at the origin in lRn. Therefore, u is bounded and harmonic in B”. Thanks to the Liouville’s theorem, we get u s 0 immediately.

6205

Let u(x) be C2 on the half-space By = {x : x, > 0) and continuous on the boundary alR; = {x : x, = 0 ) , and let g(x) be a compactly supported, C1 function on alR: = {x : x, = 0). If u, bounded, satisfies

show that its “tangential” derivatives e, j = l , - - - , n - 1, are bounded in magnitude by maxlVgl on all of BT = {x : xn > 0). (Warning: you must justify any assumptions of regularity you make on the solution u.)

(Indiana) Solution.

By the uniqueness of the bounded solution to the Dirichlet problem (3), we can show that the unique bounded solution to the problem (3) is given by

where X’ = ( ~ 1 , * - * , ~ ~ - 1 ) , I‘ = ([1,-**,(,-1) and w, is the surface measure of the unit sphere in B”.

It is not difficult to verify from the above formula that

and

476

Therefore. we have

which are just we want to show.

6206

Show that the problem

Au = -1

u = 0 for 1x1 < 1, IyI < 1,

for 1x1 = 1, au au ax ay = 0 for IyI = 1 _.--

has at most one solution. (Here u = u(x, y) is a function of two variables x and y, and is a classical solution).


We need only prove that the problem

Au = 0 for 1x1 < 1, IyI < 1, u = 0 for 1x1 = 1, au au ax ay = 0 for IyI = 1 ---

has the unique solution u 0. It is easy to see that

uAudxdy = 0. lXl < 1, lY I < 1

Integrating the above integral by parts, we have

uAudxdy

477

= -(u2(1,1) 1 - u2(-1,1) - uy1, -1) + u2(-1, -1)) 2

Then we get u f 0 immediately.

6207

Let R be a bounded normal domain in lR3, and suppose CY is such that a nontrivial solution exists to

-V2u = a 2 u in R, { u = O on a R .

(a) Show that a is a real number, and that u can be chosen to be real-

(b) Suppose a1 # a2 are such that nontrivial real solutions u1 and u2 valued.

satisfy (*). Show that r


(a) Multiplying the equation in the problem (*) by the complex conjugate of the solution u, and integrating the resulting equation in R, we obtain

(Vu12dw = a2 IuI2dv,

which implies that a is a real number. Then it is clear that u can be chosen to be real-valued.

478

(b) Suppose that u1 and u2 are chosen to be real-valued. Multiplying the equation satisfied by u1

-V2u1 = a9u1 in R

and the equation satisfied by u2

-v2u2 = aiu2 in

by u2 and u1, respectively, and integrating the resulting equations in R, we can obtain

Vul . Vu2dv = a:

and

From the above integral equalities, we get immediately

which implies the conclusion of the problem.

6208

Let fl c B“ be an open set. Let u be a continuous function on a. Prove

(i) For every x E R and r > 0 such that the ball B(x, r ) centered at x with the equivalance of these two notions of “subharmonic”.

radius r satisfies B(z , r ) cc R one has

(Here un denotes the surface area of the unit sphere in En and B ( z , r ) cc R means that the closure of B is contained within R.)

(ii) For every ball B cc fl and every harmonic function h E Co(B) such that u 5 h on dB, one has u 5 h in B.

Solution.

show that u can not take its maximum in a unless u is a constant.

(Indiunu)

Let u be subhamonic in the sense of (i). By a standard argument, we can

479

Let h be the function given in (ii). Then w = u - h is also subharmonic in

Conversely, let u be subhamonic in the sense of (ii). We define the function the sense of (i). Hence that

h as follows:

5 0 on 8B implies that w 5 0 in B.

Ah = 0 in B ( z , r ) , { h = u on aB(z ,r) .

From the definition of subhamonicity in the sense of (ii), and the mean-value theorem, we have


6209

Let R E B" be an open set, and let u be harmonic in R and continuous in

a) Show that 2 is harmonic in R for each i. b) Let B c a, where B is the open ball centered at x of radius r. Show

- R.

n that IV4z)l L ; SUP~lU(Y)l : 1% - YI = 7-1.

c) Show that, if R is bounded and 3: E R, n

lVU(.)l L -SUP{lU(Y)l : Y E 0) 4 s )

where d ( z ) = dist(z, 80).

Solution. (Indiana)

a) The conclusion is clear. b) Applying the mean-value theorem for the harmonic function in the

ball B , we have au n -(z) = - d X i wn r n

where w, is the surface area of the unit sphere in En, hence w, /n is the volume of the unit ball.

By using the Green's formula, the above equality can be rewritten as

480

And therefore, we have

c) The conclusion can be obtained from the result of the part b) immediately.

6210

Let R C lR3 be bounded and open with smooth boundary. Let u E C1(n)n C2(R) solve

A u + k 2 u = 0 in R (k > 0).

Derive an appropriate mean-value property for the solution.

Solution.

Then the spherically symmetric solutions depending only on T satisfy

(Indiana)

First we look for the fundamental solutions of the equation. Let T = 1x1.

d2 dr2 -(TU) + k2TU = 0.

For this ordinary differential equation, there are two linearly independent solutions:

u = - cos(kr), - sin(kr). 1 1 T r

Let u be a solution to the equation Au + k2u = 0. By the fundamental solution & cos k r , it is not difficult to verify that

where T,O, = (z - 2'1.

Let R < dist(zO, ds2) be a fixed positive constsant. By B(zo , R) we denote the ball centered at zo with radius R. Applying the above integral formula in

481

the domain B(zo, R), we get

Now we look for a function g(z, zo), which satisfies

(A, + k2)g(z, zo) = 0, z E B(zo , R),

cos(kT,o,), 2 E aB(zO, R). g(z,zO) = - 1 47Frxo,

It is easy to see that the function

sin( kr,o,) rxox

g(z, z0) = cot(kR)

satisfies the above conditions. For this function, it is easy to verify that

Subtracting the equation (2) from the equation (l), we obtain

a u(x0) =

1 T x o x

- cot(kR)- sin(kr,o,

- - J u(z)dS,. 4.rrRsin(kR) BB(xo,R)

This is a mean-value theorem for equation A u + k2u = 0.

6211

Let u E c2(.R2) satisfy AU = u + 1.

Prove that its spherical mean, defined by

482

and v(0) = u(O), also satisfies ( l ) , for z # 0. Here, B(0,r) denotes the ball of radius T , dSy the element of arc length.

Solution. (Indiana)

Set T = 121. Then we have

u( rw)dS, 1 W ( T ) = -

21r LB(0 , l )

and

Applying Green's formula, we get

Then it follows that

= r(w(r) + 1).

Therefore

The proof is completed.

6212

Let B = {(z,y) I x2 + y2 < 1). Prove that the problem

1 in B, u = O o n a B Au= Jm

483

has exactly one weak solution u in HA(B). (Iowa)

Solution. Let

X 1 and f =

v = d-3 w = & 5 q j 5 &G-p' It is easy to see that v , w E L2(B) and

a v a w f = - + - ax dy

in the sense of distributions (see the solution to the problem 6105). This implies that f E H-' (B) . Hence the problem admits a unique weak solution u E Hip).

6213

(a) Let R be a bounded domain in lRn with smooth boundary aQ, and f E L 2 ( f i ) , and g E L2(dR) be given. Consider the problem

gvda vv E w'q(n) (1) J , fvdx + 1, L(gradu) . (gradv) + Xuvdx =

where X > 0. Show that the problem (1) has a unique solution u E W1!2(R). (b) Will the conclusion of part (a) be valid if X = O? Explain. (c) Soppose that f , g and 80 are sufficiently smooth, write problem (1) in

( Cincinnati) the classical form.

Solution. (a) Define the inner product in W172(f2) by

(u, v)1 = h[(gradu) - (gradv) + Xuvldz, Vu, v E W'y2((n).

And denote the corresponding norm in W't2((n) by 11 - ((1. Then by applying Cauchy inequality and the trace theorem in Sobolev's spaces, we get

484

where C is a constant. Therefore,

F ( v ) = fvdx + l n g v d o

is a bounded linear functional in W'?'((n). Thanks to the Riesz theorem, there exists a function u E W'?'(R) such that

F ( v ) = (u, V)', v v E W'J(R).

Then u is a solution to the problem (1). The uniqueness is clear.

of the part (a) is false if X = 0 even for smooth f and g. (b) By the classical form of the problem, it is easy to see that the conclusion

(c) Suppose that f , g and dR are sufficiently smooth. Integrating by parts,

Therefore, the classical form of the problem (1) is

-Au + Xu = f in R, dU - = g on dR. an

6214

Let R be a unit ball in IR" centered at zero. Consider the following problem Find u E Ht(R) such that

l V u . V q 5 d x = J W X ' , V#€H,1(R), nn {X ,, =o}

where IR" 3 2 = (21, solution of that problem.

regular enough. Show that Au* = 0 and that

a , zn-l ,zn) = (x', x,). Prove the existence of a unique

Denote u+ = u Inn~X,,,o} and u- = u Inn(z,<o}, and assume that U* are

du- du+ - 1 on R n {zn = 0). ax, azn

( Cincinnati)

485

Solution. Define the scalar product in Hi(R) by

By the Cauchy inequality and the trace theory in Sobolev spaces, we can get

I: Cll4ll1,

where I( - 111 is the norm in Hi(R). According to the Riesz theorem, there exists u E Hi(R) such that

(%4)1 = / M x ' , v4 E m Q ) . nn { I ,, = o}

This proves the existence of the solution to the problem. The uniqueness of the solution to the problem is clear. Assume that u* are regular enough. Write the equation in the following

form

J,n{X.=o} 4 d ~ ' , 1, Vus + V 4 d x + L- Vu- - V 4 d x =

where R+ = R n {z, > 0) and R- = R n { x , < 0) . Integrating the above equation by parts, we have

- l+ Au+ - 4dx - l- Au- .4dx

+ 1 nn { xn=o} a x , a x , nn { x , =o} (E - ""'> dx' = / 4dx', V 4 E H ; ( Q ) .

From the above equation, we get immediately that

AU* = o in a&, au- au+ a x , a x ,

= 1 on Q n {x,, = 0).


486

6215

VU - V 4 d ~ 2 0 V 4 E W,’”(R), 4 2 0. J, Show that u 2 0 a.e. in R.

b) Let u E W1?’((n) satisfy (I) above, show that

inf u 2 inf u n an

(essinf)


a) This is a corollary of the conclusion in b) of this problem. b) If inf u = -00, the proposition is true. Assume that inf u = I > -00. Let

an

an

d(z) = max{l - u, 0).

Then it can be verified that 4 E W,’”(R) and

Now we have Vu V4dx = - IV412dx 5 0. J,

From the above inequality and (I), it holds that

IV+(2dx = 0.

By the PoincarC inequality with the above estimate, we obtain

Hence 4 = 0 a.e. in R

This implies that u 2 I a.e. in R.

487

6216

Consider functions u,, w,, w E W292(R) n W,'72((n), such that

-Awn +u, = w, -Au, - nw, = 0,

in R. Prove that '11, -+ w strongly in L2(R), as n + +oo.

Solution.

spectively, and then integrating the resulting equations in R, we get

( Cincinnati)

Multiplying the first equation and the second equation by u, and w, re-

By using Green's formula and noting that u,,v, E W,'12(R), both equations above can be written as

Vu, - Vw,dx - n11w,11i2 = 0.

Then we have

Therefore, we obtain {u,} is bounded in L2(R)

and w, -+ o strongly in ~ ~ ( 0 ) .

Multiplying the first equation by v, and integrating it on R, we have

488

By noting (1) and (2), it follows from above equality that

Multiplying the first equation and the second equation by Au, and Av, respectively, and then integrating the resulting equations, we can get

Av,Au,dx - IIVu,Il$ = LwAu,dx, -J, - J, AunAvndx + nllVun11i2 = 0.

Hence IIVUn11ia + nllVvn11iz = k VW . Vundx.

This implies that {un} is bounded in W,12(s2). (4)

Multiply the first equation by Av,. By similar consideration, we can get finally

- ~ ~ A v , ~ ~ ~ ~ - Vu, - Vv,dx = Vw Vv,dx. J, By noting (3) and (4), it follows from the above equality that

Av, + 0 strongly in L’(R).

This gives us the conclusion of the problem.

6217

Assume R c Rn is a bounded open set with smooth boundary and let f E cF(R). Suppose uk E cm(n) satisfies Auk +uk = f and has the property that for some positive number M,

uk(x)2dx 5 M

for k = 1,2,. . .. Prove that there exist a function u E C-(R) satisfying Au + u = f and a subsequence {ukj} such that ukj converges uniformly to u on each compact subset of R.

(Indiana)

489

Solution. As { u k } is bounded in L2(n), there exists a subsequence { u k j } such that

u k j converges to u weakly in L 2 ( R ) , and therefore in D'(R). Then Au + u = f in the sense of the distribution. By the regularity theorem of the elliptic equations, we have u E Cm(R).

Now we show the second conclusion. Set V k = u k - u1. It is clear that v k

satisfies A V k + V k = 0 in R

and

Hereafter, by C we denote various constants. Let R l be a subdomain of R such that 21 CC R and R = dist(dQR1)

fixed. For any z E 21, by B(z , r ) we denote the ball with the radius T (< R) centered at z. By the mean-value theorem for the equation Au + u = 0 (for the case n = 3, see the solution to the problem 6210), we have

h ( T ) v k ( z ) = 1 v k ( Y ) d S y , a B ( X , r )

where h(r) is a continuous function of T . Integrating the above equality from 0 to R with respect to T , we find that

Here R can be chosen so small that 1: h(r)dr # 0. Therefore, it holds that

I v k ( z ) I 5 c, v x E 0 1 , k = 1 , 2 , " ' . (1)

Set v k = w k + z k , where W k and z k are defined by

and

respectively. By applying the solution formula of the Dirichlet's problem for the Poisson's equation, we can get from the estimation (1)

lVWk(Z)( <_ c, vx € ill, k = 1,2, * - a .

It is clear that Igix 1Zk(.)1 5 c, V k = 1,2,. * . . a1

And by applying the mean-value property for the harmonic solutions, we can obtain from the above estimation

IVzk(x)) 5 C in any compact subset of R1.

Then by using the Ascoli- Arzela theorem, we prove the second conclusion of the problem immediately.

6218

Let R be a bounded Lipschits domain in B", and

A = {v E Hi(R) : hl 5 v 5 h2 a.e. in R},

where hl, h2 : R + .lR. are given smooth functions. (a) Show that if A # 8, there exists u E A satisfying

I(u) = minI(v). wed

(b) Show that

L V U V ( W - u)dx 2 0 VV E A.

(Iowa) Solution.

(a) Let {un} be the minimizing sequence of I(w) in A, i.e.,

lim I ( un ) = inf I(w). n-+w V E d

Define the norm of w in Hi(R) as

491

Then we get the boundedness of {IIunlll}. Therefore, there exists a subsequence of {u,}, for simplicity we still denote it by {u,}, such that

u, -u in H,~(R)

and u, ---t u in L ~ ( R ) .

And there exists a subsequence of {u,}, denoted still by {u,}, such that

u, + u a.e. in R.

Therefore, we get u E A and that

by the weak lower semicontinuity of IIwII1. This implies that

I(u) = min I ( w ) . wed

The conclusion of (a) is proved. (b) It is easy to verify that

u + t ( v - u) E A, vw E A, 0 5 t 5 1.

Let r F ( t ) = J, V(u + t ( w - u)) * V(u + t(w - u))dz.

Then F ( t ) has the minimum at t = 0. Therefore we have

F’(0) 2 0.

This implies r Vu - V(v - u)dx 2 0.


6219

Let R c IR”, n > 1, be a bounded domain with smooth boundary. Let u E C1@) be harmonic in R.

492

a. Prove that m-ax IVuJ = rnax (Vu(. n an

(Note: In both parts a and b of this question, you may cite, without proof, any standard maximum principles you are using.)

b. Suppose there exist functions @I, @2 E C2(Q) n C1(t) such that

A@I 5 0, AQ2 2 0, @I = u = on dQ.

m3x l v u l s rnax rnax lVaI1, rnax l ~ m 2 l ) . Prove that

n ( an an

(Indiana) Solution.

a. The conclusion is an immediate corollary of the following inequality: n

i , j = l

b. For the function @ 2 - u, we have

A(Q2 - u) 2 0 in R

and @ 2 - u = 0 on 8Q.

From the maximum principle, it holds that

sup(a.2 - u) = rnax(a2 - u) = 0, n an

i.e., %(.) 5 u(z ) ,

u ( z ) 5 @I(.),

vz E s2.

vx E s2% Similarly, we have

Therefore, it holds that

u(x) - u(z0) 1% - 20) I @ 2 ( 2 ) - @2(z0)

) x - 201

This inequality implies the conclusion of the part b.

493

6220

Let R be an open subset of Rn. Suppose u E C2((n) is a solution of the equation Au = u3 with the property that lVu(z)I 5 1 for each z E dR. Prove that IVu(z)I 5 1 for all x E R.

Solution. Set

(Indiana)

w = IVu12.

It is easy to see that w E C2(R) n C'((n) and

n

Aw = 2 u2izj + ~ u ~ I V U ~ ~ 2 0. 3,j=l

Therefore, w takes its maximum on (n at some point on do. Hence w 5 1 on dR implies that w 5 1 in R.

6221

Consider (a nonlinear) equation

-Au = Xu2( 1 - u) in R,

u = 0 on dR, (1)

where X > 0 is a parameter, and dR sufficiently smooth. (a) Prove that for any solution u of (l), 0 5 u 5 1. (b) Show that if X > 0 is sufficiently small, the only solution of (1) is u 0.


at to and (a) If u 2 1 is false, then there exists xo E R such that u takes maximum

u ( 2 ) > 1.

For the maximum point xo, it is clear that

Au(zo) 5 0,

i.e., -Au(z') 2 0.

494

This contradicts the assumption that

x u y 1 - U)(ZO) < 0.

Hence u 5 1. As Xu2(1 - u) 2 0, u can not take minimum in R unless u f 0. Hence

(b) Multiplying the equation by u and integrating the resulting equality on u 2 0.

R, we can get

J, 1vul2dz = x u3(1 - J, By using the conclusion of (a), it holds that

Applying PoincarC inequality, we obtain from the above inequality

where C is a constant. If X < C-l, we have u E 0 immediately.

8222

Let R be a bounded normal domain in IRn with smooth boundary dQ. Let ?i' be the exterior unit normal on dQ. Show that the only solution of the boundary value problem for the biharmonic equation:

A(Au)=O i n Q

u = 0 on dR,

= 0 on dR, dU dn -

is the trivial one u E 0. Assume u E C4((n).

Solution.

have

(Indiana-Purdue)

Integrating by parts and taking the boundary conditions into account, we

r 0 = JnA(Au)uda

495

= l A ~ 1 ~ d x .

Therefore, u is harmonic in f2. Taking into account u = 0 on af2, we get u immediately.

0

496

SECTION 3 PARABOLIC EQUATIONS

6301

Consider the Cauchy problem for the Heat Operator in IR1:

ut = u,, (-0O < 2 < 0O,t > 0)

u(2,O) = f(2) (-0O < 2 < m),

where f is bounded, continuous, and satisfies

03 J_, lf(”)I2d2 < 0O.

Show that there exists a constant C such that

for all -00 < 2 < 00, t > 0.

Solution. (Indiana)

The solution to the problem is given by the Poisson’s formula

u(2,t) = - Jrn f ( y ) e - W d y . 2m -m

By the Cauchy inequality, we have

Set y = 2 + &q in the second integral of the above inequality. Then we get immediately

c lu(2,t)l I t1/4‘

497

6302

Let R c Bn be an open set with smooth boundary and suppose u E Cm(R x [0, m)) is a solution of the equation

with u = 0 on a R x [ O , 00). Assume that

Let $(t) = 1 u(z,t)2dz.

n Prove that $(t) + 0 as t 4 +m.

Solution.

on R, we can get

(Indiana)

Multiplying the equation by u and then integrating the resulting equation

The PoincarC inequality gives

IIU(*, t)ll:z(n) 5 CIIVU(*, t)l&(n)?

where C > 0 is a constant. For any E > 0, it holds that

Taking E = 1 /C , we obtain

Solving this differential inequality, we have


if l/f[-,t)ll&n) .--) 0 as t + +m. This completes the proof.

6303

Consider the Cauchy problem

ut = u,, (-00 < 2 < 00,t > O ) ,

4 2 , 0) = f(z),

where f is bounded and continuous.

for this problem, and write down the solution u.

non-uniqueness example possible?

a) Using the Fourier Transform, construct explicitly a fundamental solution

b) State a maximum principle for this problem. Why is the Tychanov

c) Suppose in addition that f E L 1 ( B ) . Show that

lim u(2, t ) = 0. t-a,

d) Show how to solve the problem

ut = u,, (2 > 0 , t > 0)

u(z,O) = f (z ) (2 > 0) U ( 0 , t ) = 0 (t > 0)

by using a) and an appropriate extension of f.

Solution.

forming the following Cauchy problem with the initial data S(z):

(Indiana)

a) Let G(<,t) be the Fourier transform of u(z , t ) with respect to z. Trans-

ut = u,, (-.. < 2 < 00,t > 0)

4 2 , 0) = 6(2),

we get

499

The solution to the transformed problem is given by

G(<, t ) = e-tzt.

Then the inverse Fourier transform of 6 gives the fundamental solution to the

From this fundamental solution, we can get the explicit representation for the solution u(z , t ) to the general Cauchy problem with the initial data f (z )

u(z,t) = - J 4 ( y ) e - W c l y . 2- n-21

b) The maximum principle can be stated i ~ s follows: Suppose that u(z, t ) is bounded and continuous on lR x [O,T], and satisfies the heat equation in lR x (O,T] with the initial data f(2). Then it holds that

inf f (z) 5 u(z , t ) 5 sup f(z) X E R X E R

for ( z , t ) E lR x (O,T].

The Techanov non-uniqueness example is possible for some unbounded so-

c) The conclusion is clear. d) Defined an odd function 4(z) as follows:

lutions.

2 > 0, f(x>, 4(.) = { -f(-z), 2 < 0.

Then by solving the following Cauchy problem:

ut = u x x (-cm < 2 < cm,t > O ) ,

42,O) = 4(z),

we get the solution to the initial-boundary problem

6304

Consider the problem

U t = u,, (z E IR,t > 0) u(2,O) = uo(z) E L 2 ( E ) .

Find a bound on 1:' Ju,(z,t)I2dz in t > 0.

Solution.

we can find that

(Indiana)

From the Poisson's formula for the Cauchy problem of the heat equation,

Then by using the Cauchy inequality, we have

And therefore, it holds that

We can also get the result by the method of energy integrals. Multiplying the heat equation by u and tut respectively, and integrating

x (0, T), we can obtain the resulting equations in

and

tu,,(z, t)ut(2, t )d zd t . I' L: tlut( 2, t ) 12dzdt = 1' 1: Adding the double of the second equality to the first one, we have

501

otherwise an approach to the function U O ( X ) by a sequence of functions in Cr(En) can be used.

From the above two equalities, we have oo

A / m ~ u ( ~ , T ) ~ ~ d z + T L _ lu,(z,T)I2dz 2 - w

Therefore, we get an estimation

6305

For the indicated domain R (interior of the parabola) let RT denote the points ( z , t ) with z E R and t < 2'. Let B1 denote the open line-segment forming the top of ~ R T , and let B2 = BRT\B~. Let u(z, t ) E C o ( n ~ ) with u,,, ut E C o ( R ~ U B1) be a solution of

%x - ut + a(2, t)u = f ( 2 , t>

502

with a, f E Co(D). Show that if a < 0, f 5 0 in RT, then every non-constant solution u assumes its negative minimum (if one exists) on B2. What can you say if a < 0, f 2 0 in RT?

(Iowa)

Fig.6.1

Solution.

we have Suppose that u assumes its negative minimum at ( ; G O , to ) E RT U B1. Then

u(zO,tO) < 0, ut(z0,t0) 5 0

and u,,(zO,tO) 2 0.

u,,(2 0 0 , t ) - ut(z0,tO) + a(zO,to)u(zo,tO) > 0. Hence

This contradicts f(zo, to) 5 0. (Here a < 0 should be assumed in RT u B1).

its positive maximum (if there exists one) on Ba. If a < 0, f 2 0 in RT, we can say that every non-constant solution assumes

6306

Consider the boundary-value problem

u,, + ut = 0 (0 5 z 5 A and t > 0), B.C. u(0,t) = 0,

UP( . . , t ) = 0, I.C. u(z,O) = f(z).

Use the sequence { f n ( z ) = 2 n + l sin ( y z ) } and an appropriate space of continuous functions to decide whether this problem is well-posed.

( Indiana-Purdue)

503

Solution. When n is even integer, f n ( z ) is the eigenfunction of the corresponding

two-point boundary value problem. By setting u = f,(z)T,(t), we can find the solution u, to the problem with the initial data f(z) = fn(z)

for all f E C[O,r]. It is clear that

But for any fixed t > 0

IIu,(-, t)ll -+ 00 as n + 00.

This implies that the problem is not well-posed in the space of continuous functions with the norm defined above.

6307

In Probability Theory one encounters the Ornstein-Uhlenbeck process, in which the particles of Brownian motion are subjected to an elastic force. One is required to solve

ut = u,, - xu, (x E az, t > 0) u(z, 0) = uo(2).

Assume uo E S and find u. (You should be able to find u explicitly; if not, leave your result in the form of a one-dimensional integral.)

Solution. (Indiana)

By G(<, t ) we denote the Fourier transform of u with respect to 2. It is easy to see that

d a - a;r; F(zu,) = i-F(u,) 2 --(<G) = -u -I-. at at at

504

Then the transformed equation and initial condition are in the following form

qt, 0) = G o ( t ) .

Solving the above Cauchy problem for the first order linear partial differential equation, we can obtain

,. u(<, t ) = et-3(e2'- l)caiio(et<).


F-'( i io(et t ) ) = e - tuo(e - t z )

and 1 - 2 2(e-- - 1) . F-l(e-li(e2t-l)€2) =

d m e Therefore, we get the solution to the problem

Remark. From the above solution formula, we can find a convenient transformation of variables to the problem. In fact, set

1 2

T = -(I - e - 2 t ) , t = e - t x .

Then the problem is transformed into the following

7 = 0 : 21 = U o ( t ) .

The above solution formula can be obtained easily from the Poisson's formula for the Cauchy problem of the heat equation.

6308

Let L be the usual heat operator

ax; a2 1 ' a L = - - -+...+- at (;I:

let 4 E D(nZ"), and let h be a fixed point of IR". Consider the problem

505

(Lu)(z , t ) = u(2 - h,t) , 2 E B",t > 0,

u(2,O) = 4(2), 2 E B"

for an unknown function u(z, t ) . Derive an explicit representation for a solution u(z , t ) of the above

problem in terms of the initial data. (You should prove that your u is well- defined, but you need not prove that it represents a smooth solution.)

b. Assume that 4 is the Fourier transform of a function which vanishes in the ball of radius R 2 0 centered at the origin. Show then that the solution u satisfies

a.

Il"(',t)llL~(R") < - e(1-R2)tl141~L2(Rn).

(Indiana) Solution.

the above problem we obtain a. Let G((, t ) be the Fourier transform of u with respect to 2. Transforming

i& + ( [ ( I 2 - e-"n€)G = 0, < E En, t > 0, 2 E ELn { O ( < , O ) = m,

which admids the following solution

qt, t ) = &-(()(y ( l€12-exP( - i h . € ) ) t .

The inverse Fourier transform of G gives us the explicit representation for the solution u(z, t ) to the problem

This is just the conclusion of the problem.

506

6309

Let R c IR" be a bounded domain and assume u ( x , t ) 2 0 is a function with u E C2(o x [O, oo)), which solves the equation

4 U+, - AU = -U

(heat conduction with heat loss due to radiation) with the boundary condition u Ian= 0. Prove that we can find a constant C such that

E(1) = u2(x , 1 ) d x 5 C I regardless of the initial value u ( z , 0).

Solution. (Iowa)

Multiplying both sides of the equation by u and integrating on f2, we have

u t u d x - Au - u d x = - u5dx . I By using the Green's formula and noting the boundary condition, we get from above euualitv

--E(t) + IVu12dz = - u 5 d z , 2 I d dt 1 1

where E ( t ) = & u 2 d x . The Hijlder inequality gives

i.e.,

Using above estimation, we get from (1)

I d 2 dt - -E( t ) 5 - J R J - + E q t ) .

507

The above inequality can be written as

Integrating the above inequality from 0 to t , we have

E-f ( t ) 2 E-+(O) + 3(R(-;t.

It follows that


6310

Let

satisfy ut + u3 + sin(u) = A,u, u = h, u = 9 ,

B(0, T ) x (0, TI, W O , T ) x LO, TI, B ( O , T ) x (0)

(In diun u) with g, h continuous and g, h < 1. Prove that maxu < 1 on B(0,r) x [O,T].

Solution.

(O,T] such that u takes the maximum a t ( zo , to) and If the conclusion of the problem is false, then there exists (zo, to) E B(0, T ) x

u(z0, t o ) 2 1.

It is clear that ut(zo,to) 2 0 and Az(zo,to) 5 0.

Therefore, we must have

But from the equation we get

(ut - A+u)(zo,tO) = -(u3 + sin(u))(zo,tO) -1 - sin 1 < 0, -u3(z0,to) -sinu(zo,to) < 0,

if u(zO,tO) = 1, ifu(zo, to) > 1.

This contradicts (1).

508

6311

Let u E C2(R x (0, T)) n Co(n x [O,T]) be a solution to the problem

~t = AU - u3 2 E R , t > 0, z E dR, t > 0:

u(z,O) = f (z ) 2 E a, { u = o

where R is a smooth bounded domain in En. (a) Prove that

IIUllLz(n)(t) 5 IlfllL.(n) for all t E (0, TI.

IIUIIL-(n)(t) L llfllL-(n) for all t E (0, TI.

(b) Prove that

(c) Prove that the solution to this boundary value problem is unique. (d) Show that for f 6 L4(R) n H1(R), the following bound also holds:

I I44(n , ( t ) + II~llLl(n)(t) 5 Ilfll;yn) + IlfllLl(n). (Indiana)

Solution.

on 0, we get (a) Multiplying the equation by u, and integrating the resulting equation

Au . udx - u4dx.

By applying Green's formula, we obtain from above integral equality

Then we get the conclusion immediately. (b) First we show that u can not take the positive maximum in R x (O,T].

In fact, if u take the positive maximum at ( zo , to) E R x (O,T], then we must have

ut(zo,to) 2 0 and Au(zo,to) 5 0.

Hence (ut - Au)(xO,to) 2 0.

This contradicts the inequality

-u3(xO,tO) < 0.

In a similar way, we can show that u can not take the negative minimum

Therefore, we have in R x (O,T].

Il.llL-(s2)(t> L llfllL-(ct), vt E (0,Tl.

(c) Let u1 and 212 be both solutions to the boundary value problem, and w = u1 - u2. The w solves the following problem:

W t = AW - (u: + ~ 1 ~ 2 + u ~ ) w , w = 0,

x E R, t > 0, x E a R , t > 0, x E $2. { w(z, 0) = 0,

By a similar deduction as done in (a), we can get

Il4lL.(ct)(t) 50, vt E ( 0 7 1

for above problem. This implies that u1 = u2, and the uniqueness of the solution is proved. (d) Multiplying the equation by ut and u3 respectively, and integrating the

resulting equations, we get

1 u t d x = J, Au - utdx - u3utdx

utu3 = J, Au . u3dx - J, u'dx.

J, and

By applying Green's formula to the first terms in the right side of the both equalities above, we can obtain

f (a L u ' d x ) + dt d 1 ( J, lVu12dx) = - l u i d x 5 0

5 dt (1 4 J, u'dx) = - 1 3 ~ ~ I V u ( ~ d x - J, u'dx <_ 0. and

Adding the both inequalities above, we have

5 dt ( / , u 4 d x + J, IVul'dx) 5 0.

510

This differential inequality, combining with (a) if necessary, implies the conclusion of (d).

6312

For each p E En, let A(p) be a real positive definite n x n matrix with entries aij(p). Assume A E C'(Bn). Let R C B" be a bounded open set. Prove (from first principles) the following comparison principle:

Let u and v be C2 solutions to the inequalities

~t 5 aij(Vu)uzizj for ( ~ , t ) E R x ( 0 , ~ )

and wt 2 aij(Vw)vzizj for ( z , t ) E R x (0, co),

respectively. Suppose, furthermore, that u and w are continuous on n x [O , co) and satisfy the condition u 5 w on the parabolic boundary

(Indiana) Solution.

Set w = u - v and

Then from the given differential inequalities, we can obtain a differential inequality for w as follow:

n n

i j= l k = l

By applying the maximum principle to the above differential inequality, we find

w 5 o on I x [0 , co).

This is the result we need.

51 1

6313

For R c Rn bounded and open, let u E C2(n x [0 , m)). Assume u i j (z , t , z , p ) and b i ( z , t , z ) are continuous functions of their arguments and assume that

n

C ~ i j ( z , t , z , p ) t i t j L x( z , t , z , p )1 t12 > o for every t E R~ - { o ) , i , j = l

where X is a positive function. Suppose u satisfies the inequality n n

ut - C ai j (z , t , u, vu)uZixj - C bi(z, t , u)uxi >_ o for z E R, t > 0. i,j=l i= l

Prove that for any T > 0

min u(z, t ) = min u(z, t ) 0 x P,TI Q -

where Q = (R x (0)) U (dR x [O,T]).

Solution. (Indiana)

First we prove that if w E C2(a x [O,m)) satisfies

wt - C a i j ( z , t , u , ~ u ) w , ~ , ~ - C b i ( t , z , u ) w Z i > 0, n n

at ( q t ) E 0 x ( O , T I ,

(1) i , j = l i = l

then ( z , t ) can not be the minimum point of w in R x (O,T]. In fact, a t the minimum point (z, t ) , it must hold that

Wt 5 0, wxi = 0

and

which contradicts (1). Now suppose that u takes the minimum at (%',to) E R x (0, TI and

u(zo, to) < min u. Q

512

Take a > 0 large enough so that

a u l l ( z , t , u, Vu) + b l ( z , t , u, Vu) > 0, for ( z , t ) E s2 x [o ,T] .

Then w = u - &eaxl

takes its minimum at some point in R x (O,T], if E > O is sufficiently small. However, for the function w we have

This is in contradiction with w having minimum point in R x (0, TI.

513

SECTION 4 HYPERBOLIC EQUATIONS

6401

Assume that g is smooth and in L 1 ( R f l ) , n arbitrary. Let G ( < , t ) be the Fourier transform (with respect to z) of the solution to the n-dimensional wave equation:

Utt - c2 AU = 0, u(2,O) = 0,

+ , O ) = 9(z) .

Show that the LP norm of .^(.,t) at time t for p > n is bounded by a constant times t l-nlp.

(Indiana) Solution.

Transforming the equation and the initial conditions, we have

2tt + c21<)22 = 0, K O ) = 0, %(t, 0) = $(tL

G(<,t) = --?( n ensures the convergence of the integral. The above estimation gives the conclusion of the problem.

6402

Consider the Cauchy problem for the wave equation

Assume U O , U ~ E S. (a) Use the Fourier-transform to show that the total energy at time t

is constant in time. (b) Let Eo denote the constant energy in (a), i.e., E( t ) Eo. Show that

, ,

(Iowa) Solution.

conditions with respect to the variable 2 = (zl,. - . , zn), we have (a) Taking the Fourier-transforms of the wave equation and of the initial

utt + l€12G(t,t) = 0, q€, 0) = Go(€), G(€, 0) = Gl(€) ,

where < = (€l,---,[n). differential equation, we get

Solving the above initial problem of the ordinary

%(€) It I qt, t> = Go(€) cos(ltlt) + - sin( I€lt)

51 5

Integrating the above equation on En and noting that

we obtain

Then the Plancherel theorem gives us

Combining this with (a), we finish the proof of (b).

6403

Consider the initial problem for the wave equation in three space variables:

u,,,, + %,,, + u,,,, - U t t = 0; 2 E B3,t > 0, u(z,O) = qqz); z E R3, U t ( 2 , O ) = $(z); z E B3.

516

(a) Write down the formula for the solution of the problem. (b) If q5 and II, vanish outside a ball of radius 3 centered at the origin, find

(c) If 4 vanishes everywhere in R3, and the set of points in lR3 where you are sure that u vanishes when t = 10.

0, for 121 < 1, { 0, for 2 < 121,

$(z) = k , for 15 1x1 5 2,

where k is a constant, find u ( 0 , t ) for all t 2 0. (Your answer should be explicit, no integrals).

(Indiana -Purdue) Solution.

(4

(b) By applying the domain of dependence for the solutions to the wave

In fact, when 121 5 7, it holds that equation, we can verify that u vanishes when t = 10 for 121 5 7 or 121 2 13.

IyI 2 3 for all y E {y : Iy - 21 = 10).

Therefore

J q5(Y)dS = J II,(Y) = 0. Iy-zI= 10 ly-zl=lO

Similarly, we can show that the above equality holds when 121 2 13. (c) It is easy to verify that

0, O < t < l ,

0, 2 < t .

6404

Let R be the upper-half space in lRn

R = {(21,22,23) E lR3; 2 3 > 0).

51 7

Consider the initial-boundary value problem

a z u a z u azu a z U ax: ax; ax; a t 2 - + - + - - - = 0. , X E R , t > O ,

U ( X , O ) = d(X), W ( X , 0) = +(x); 2 E R, E aa, t 2 0. t ) = 0,

(a) Write down a formula for the value u(x , t ) of the solution at ( x , t ) , x E R, when

t < 2 3 .

(b) Find a formula for the value u(z, t ) of the solution at (2, t ) , x E R for all t > 0.


(a) When x E R and 0 < t < 2 3 , the domain of dependence for the value u(x, t ) at (2, t ) is in R. The value u(x, t ) is given by the formula for the Cauchy problem of the equation:

(b) Define *(x) and @(x) as both odd functions with respect to 2 3 in E3 as follows:

4 ( 2 1 , z 2 , 2 3 ) 7 2 3 2 0, -4(x1,m, -x3) , 2 3 < 0,

--+(% 2 2 , -231, 2 3 < 0, ‘$(xl, 2 2 , 2 3 ) , 2 3 2 0, @(z) =

respectively. Then the value u(x, t ) is given by

If both 4 and II, satisfy certain compatibility conditions, it is easy to see that u(x, t ) given by the above formula satisfies the wave equation and the initial conditions. When 2 3 = 0,

J y . = ~ t ~ - ~ y l - x * ~ ~ - ( y l - x ~ ~ ~ + ( Y l , Y2, Y 3 W

-/ Ys=-~/ta-(Yl-zl)a-(Yz-za)a +(Yl,Y2, -Y3)dS = 0.

518

In a similar way, we have

@(y)ds = 0, when 23 = 0. J Jy-xl=t

Therefore, u(z,t) satisfies the boundary condition on 23 = 0.

6405

Let u(2,t) be a solution of class C2 to the Cauchy problem

with compact support. Define the local energy by

(i) Use the energy indentity to show that

Conclude that the total energy EW(u(t)) is conserved. (ii) When n = 3, evaluate lim E ~ ( u ( t ) ) . (iii) Do (ii) above when n = 1. (iv) Let 7~ = 2, #

t+O"

0 and suppose that $ 0 for 121 2 k, where 2: = ( ~ 1 ~ 2 2 ) . Show that u satisfies

for some constant c, on the set T < t - 4k. (Here T = 1.1).

Solution. (Indiana)

(i) For t E [O,T], we have

E R - T + t ( u ( t ) ) = (u," + IVuj2)dSdp.

51 9

By the above expression, we can calculate that

Hence ER-T(U(0)) 5 ER-T+t(U(t)) 5 ER(u(t))-

Similarly, we can show that for t E [O,T]

It is clear that the total energy E,(u(t)) equals to E,(u(O)). (ii) From the formula of the solution to the problem

it is easy to verify E ~ ( u ( t ) ) = 0 if t > R + a,

520

where a is the radius of a ball in which supp4 U supp.ll, is contained. Hence

(iii) By the formula

it is easy to evalute lim ER(u(t)) = 0.

t-oo

(iv) When n = 2, the solution is given by

If 1x1 < t - 4k, it is easy to see that

Hence the formula of the solution can be rewritten as

When 1x1 < t - 4k and IyI I k , we have

IY - 4 I 14 + 2 k ,

and hence 1 1

JW I J(t + IzI + 2 k ) ( t - IzI - 2 k ) ' It is easy to verify that

< d3 Vt - 1x1 > 4k. t - 1x1 + 2 k .J t - I z l - 2 k -


521

6406

Let R c B" be a bounded domain and assume u(z , t ) is a function with u E C2(a x [0, oo)), which solves the equation

with the boundary condition = 0. With

prove that there is a C < 00 such that

for all t 2 0.

Solution. (Iowa)

Multiplying the equation by ut and integrating the resulting equation with respect to x in 0, we

Here we have applied

have

the Green's formula:

By the Poincar6 inequality, we can conclude that

Thus we obtain a differential inequality for E(t )

d dt -E(t) I CE( t ) .

Then E( t ) I exp(Ct)E(O) is an immediate consequence of the above differential inequality.

522

6407

(a) Find the explicit form of the solution of the initial value problem

v2u = U t t , 2 E B3,t > 0,

u(z,O) = fo, z E B3, r

u,(z,O) = -m, r 2 E B3,

where T’ = z:+zg+zi, and f(q) is a C3 function on the real line with compact support such that

where f’(q) is the derivative of f with respect to q. (b) What is its energy at time t = 10.

(Indiana -Purdue)

Find the spherically symmetric solution to the problem. In the symmetric Solution.

case, the equation can be written as

a Z u 2 a u - + -- = U t t . ar2 r ar

Set w = TU. Then w solves the following problem:

w,, = wtt, w = 0,

T > 0,t > 0, r = 0, { w = f ( r ) , wt = -f’(r), t = 0.

Solving this problem, and then we can get

(b) The energy of the problem is given by

1 E ( t ) = - (ut” + IVul2)dz. 2 L.

It is easy to verify that

E(t ) = E(O), vt > 0.

523

And therefore

= 27F

6408

Consider the wave equation in R3:

utt - Au = 0 for x E B 3 , t > 0,

u(z,O) = 0 ,

U t ( 2 , O ) = g(z),

where g E Cr(R3). Prove that there exists a constant C depending only on the given data such that


Let R and M be positive constants such that

and Ig(z)l 5 M , v x E R3.

From the Poisson’s formula, we have

It is easy to verify that

The area of the intersection {y E R3, Iy - X I = t } n BR 5 The area of aBR.

524


MR2 t > 0. 1

lu(z,t)l 5 - . M .4rR2 = - 47rt t ’

6409

Let u E C2(.Rn+l(x, t ) ) be a solution of the equation utt -Au = 0. Suppose also that u = ut = 0 on the ball B = {(z,O) : 1x1 5 t o } in the plane t = 0. Prove that u vanishes in the conical region

D = {(x,t) : 0 5 t 5 t o and 1x1 5 t o - t } .

(Indiana) Solution.

Let = {(Z,t) E En+1, 121 5 t o - t , t = T }

for 0 5 r 5 to. Multiplying the wave equation by ut, and integrating the resulting equation with respect to x in R,, we can obtain

Set 1

E(7) = 5 l r ( u , ” + IVuI2)(x, T ) ~ z .


Therefore, we have

525

It is clear that E(0) = 0. Hence

E ( 7 ) = 0, vo 5 7- 5 to.

This implies that u=O in D.

6410

Let U(Z, t ) be a solution of the Cauchy problem

utt = AU (Z E B3, t > 0 ) ,

u(z, 0) = f(z), Ut(Z,O) = d x ) .

Assume that f , g E Cr(nZ3). Prove that if /ltnL8u 2 dx = 0,

then

Solution. Let R > 0 such that

gdx = 0. I.. (Indiana)


for any t > R. By using Green's formula, we get the following equality from the equation

Audx = 0. J,, uttdx = J., Therefore, we have

526

and

by the initial conditions. In the other side, by Cauchy inequality it follows that

21ax 5 (J,.iax)t (/ ax) l I 2 J R S t- R< (3Cl <t+R

for any t > R. This contradicts (1) provided that

J,, gdx # 0.

6411

Consider the solution of the wave equation

U t t = nu (x E R3,t > 0)

with data u(x, 0) = 0, ut(x, 0) = +(x) E Cp. a) Use Fourier transform to show that there is a positive constant c such

that JlzIRs u 2 dx = c.

b) Show that there is a positive constant C1 such that

t - m a x l u l ? CI X

for all sufficiently large t.

Solution. a) Let 6 denote the Fourier transform of u with respect to the variable

x = (XI,. . . , zn). By taking the Fourier transforms of the wave equation and the initial conditions, we can get

(Indiana)

527

Therefore

we have

Then we get

where

if t+b $0 . b) Let Q(t) be the support of u ( z , t ) with respect to the variable 2 =

(21,21,z3) for t > 0. If the conclusion is false, then there exists a sequence { i n } such that

t , 4 0 0 , a s n - + c c

t, - m a x ~ u ( z , t , ) ~ < la(t,>l-+.

Iu(2,tn)l I IQ(t,)l-+t,’.

and

2

This gives

Then we have

This contradicts (a). The proof of (b) is completed.

528

6412

Consider the Cauchy problem for the wave equation

a) Let n = 3; q5,11, E CF(R3). Let 8 < 1. Determine the large time behavior of the integral

J u4(2,t)dz. l4<$t

b) Let n = 2, 4 3 0, 11, E C F ( R 2 ) . Show that for any 19 < 1, we have

sup Iu(z,t)l = O(t-') I x l 9 t

as t -+ oc).

(Indiana) Solution.

a) Assume that R > 0 is so large that

suppq5 u supp11, c (2 f IR3, (21 < R}.

From the Huygens's principle for the %dimensional wave eauation, we conclude that if t > R and

121 < t - R,

then u(2,t) = 0.

Therefore, when t > A, it holds that

+, t ) = 0, vlzl< et,

which implies that

R u4(2, t )dz = 0, when t > __ J lxl<@t 1 - 8 '

b) From the Poisson's formula, the solution to the 2-dimensional problem is given by

u(21, 2 2 , t ) = -

529

Let R > 0 be so large that

suppllr C {z E I R 2 , 121 < R } .

When t > 5, it is clear that if IzJ 5 B t and Iyl 5 R then

JY - X I I et + R

and dt2 - (y - zI2 2 dt2 - ( B t + R)2.

Therefore, we have

where C is a constant. The proof is completed.

6413

The linear transport equation is following PDE for a scalar function f of seven variables z, v, t (z E B3, v E IR3, t E R;):

ft + v . V x f = 0.

Consider the initial-value problem for this; i.e., let

f ( c , 'u, U ) =j (c, v); j given, j~ ~ 1 .

0

a) Show that, if f > 0 for all z, v, then the same is true for f at all later

b) Define the local density p by times .

p(.,t) = J,, f ( Z , v, t)dv-

0

Suppose that 0 5 f (z,v) 5 h(z), where h E L1(B3). Show that

supp = o(t-3) as t + oo. X

(Indiana)

Solution. a) The characteristic curves of the equation are defined by

d x dt

= v -

and along which it holds that df - = 0. dt

Therefore, the forward characteristic curve departing from t = 0, x = a is

x = (Y + tv. (2)

From (1) and (2), we get the solution to the initial-value problem

0

f (2, v,t) =f (x - tv, v). 0

It is clear that f 2 0 i f f ? 0. b) From (3), we have

f (x - tv, v)dv = JR3 O


6414

Solve the Cauchy problem

xu, + yuu, = -xy ( x > 0) u = 5 o n x y = 1,

if a solution exists.

Solution. ( h dian a)

The system of the characteristic equations for the problem is given by

531

du = -xy _ - dx dY ds --, z = y u , - ds

with the initial conditions

1 s = O : x = a , y = - , u = 5 .

(Y

Fkom the above Cauchy problem, we have immediately

and du du d ds ds ds 21- = - - ~ e ' - = _- (ae'y + u).

-u2 + u + x y = c, This implies that

1 2

where C is a constant. The initial data gives C = F. Then we obtain immediately

u = -1 + 43cxy.

6415

Compute an explicit solution u(x, t ) to the initial-value problem

Ut + = 0, U(X,O) = z2.

(Indiana) Solution.

Set v = u,. Differentiating the equation and the initial condition with respect to 2, we get

vt + 22121, = 0,

W ( 2 , O ) = 22.

The characteristic curve of the quasilinear equation (1) is given by

dx - = 2v, dt

532

along which it holds that dv dt

= 0. -

From the above characteristic equations, we can obtain immediately 22 v = -

4t + 1'

22

4t + 1'

Hence u= -

6416

Solve the following initial value problem.

ut + uus = 0, 2 E IR,t > 0,

u(z,O) = 1 - 2 , 0 5 z < 1, 2 < 0, {:: 2>1.

Show that the continuous solution exists only for a finite time, and find the discontinuous entropy solution, giving explicitly the discontinuity curve and the Rankine-Hugoniot jump condition along it.


The characteristic curve departing from t = 0, z = P is given by d x =u, t = o : x = p , dt __

along which u is a constant, i.e.,

d u _ - - 0 , dt

t = 0 : u = u"(p),

where uo(z) = u(x,O). From (1) and (2), we have

2 = uO(P)t + p, u = uO(P), and

P < 0,

P > 1, 1, P < 0,

0, p > 1. 21 = { 1-p , o 1,

for 0 5 t < 1 (see Fig.6.2).

Fig.6.2

When t = 1, the continuous solution to the problem blows up. For t 2 1, the Rankine-Hugoniot condition along the discontinuous curve

is given by

dt i.e.,

dx 1 dt 2 - = -(u+ + u-).

Noting u+ = 0, u- = 1, we have

dx 1 dt - 2 ' _ - -

Therefore, the discontinuous curve is

1 2

x = - ( t + 1 ) , t l l

534

and the entropy solution to the problem is given by

2 < ' ( t + l), 3 2 > z( t + 1). 0,

6417

a) Let u(z , t ) be a smooth solution of

at4 a U -+a(z,u)--bu=O f o r O < t < T , at d X

where b is a constant and a is smooth. Show that if

a 5 u(z , 0) 5 p for all X,

then aebt 5 u(z , t ) 5 pebt

for all x and all t E (0, T) . b) Let u(X,t) be smooth solution of

d U 8U - + a(u)- = 0 at d X

for 0 < t < T,

U ( X , O ) = U O ( X ) ,

where uo and a are smooth. Show that, if z ( t , y ) is the characteristic through (Y, O), then

U o ( Y , 0) ( & + h 3 ( z ) ) l z = y ) t + 1'

u z ( 4 t , Y), t ) =

i) Use this to compute the largest T for which a smooth solution can exist

ii) Show that, if 0 5 t l < t2 < T , then up to time T.

00 00

(Indiana) Solution.

a) Let u = ebtv. Then v satisfies the following equation

bt d V - + U ( X , e v)- = 0. at d X

535

The solution v maintains a constant along a characteristic of the above equation, defined by the equation

dx dt - = a(x, I?”).

Therefore, it holds that

infw(x,O) 5 v(z,t) 5 supv(z,O). X 2

This gives the estimation in the problem immediately. b) u(x, t ) is a constant along the characteristic z = x(t, y). Hence

Differertiating above equality with respect t o y, we have

The characteristic x(t, y) through (y, 0) is given by the following Cauchy problem:

&, Y) = a(uo(y)), { a 40,Y) = Y.

Differentiating this problem with respect to y, we get

g (g+, Y)) = a’(uo(y))ub(y), { a FZ(0,Y) = 1.

Then it follows from above problem that

Therefore, we have

This is just that we want to show. i) Let m = inf a‘(uo(y))ub(y). From the above expression, we can get the

If m >_ 0, the Cauchy problem admits a global smooth solution in ( 0 , ~ ) ;

Y following conclusions easily.

536

If -co < m < 0, the largest T for which the problem admits a smooth solution in (0, T) is given by

mT+ 1 = 0.

ii) For any fixed t E (0 , T), by the change of variables z = z(t , y) in the integral, we have

00

Taking the expressions of u,(z(t, y), t ) and &z(t, y) into account, we obtain

00 00

This completes the proof of the problem.

0418

(Uniqueness of weak solutions) Consider the Cauchy problem

'Zlt + '11, = 0, u(z,O) = uo(z),

on IR x [O,T], on IR x (0).

A function u E P ( I R x [O,T]) is defined to be a weak solution of (2) iff

lT /: 4 Y , . ) ( -4t(Y, - 44YI s))dyds

+ J_, .uO(Y)4(Y? O P Y = 0 00

(2')

for all test functions 4(z , t ) E P ( I R x [O,T]) which are compactly supported and vanish on {t = 7').

If uo(z) E 0, prove that the only LP solution of (2') is u G 0. (Indiana)

Solution. Consider the following Cauchy problem:

4t + 4, = f ( t , z), on x [0,7'], 4(.,T) = 0, on IR x {T},

537

where f E Cm(B x [0, T]) with a compact support.

~5 E Cm(E x [O,T]) which is compactly supported. Therefore, we have It is not difficult to show that the above Cauchy problem admits a solution

6’ J_+_m U(Y, .)f(Y,5)dYd5 = 0 (*I

for all function f E Cm(B x [O,T]) with a compact support, provided uo(x) 0. As the space Cr(B2 x [0, TI), consisting of all Cm-functions with compact support in E x [O,T], is dense in L * ( B x [O,T]), where q = p / ( p - l), the equality (*) is valid for all functions f E L q ( B x [0 , TI). Hence we have u 0.

6419

Consider the Cauchy problem for a function u = u(x, t ) , where x E B and t > 0.

ut + a(z,t)u, = b(x,t), u(z, 0) = uo(x) E C(B2).

a) Show that if b TV = “total variation”).

b) Show that when b $ 0, one still has the bound

0, then TV(u(., t ) ) 5 TV(uo(.)) for each t > 0 (where

TV(u(* , t)) 5 TV(u0) + TV(b(*, 5))ds.

(Indiana)

J,’ Solution.

The characterstic curves of the equation are given by the equation

dx dt - = u(2, t),

along which the solutions to the equation satisfy

du dt - = b(x, t ) .

a) Suppose that b 0. Then u is a constant along a characteristic curve. We divide the line t = T into subintervals by the points * * , x - ~ , . . * , 2-1, zo, zl,. . . , z,,, - . .. Suppose that the backward characteristic curve through (xi, T) meets the line t = 0 at ( 0.

b) Let { z i } and {ti} be given as in a). When b $ 0, we have

u ( z i , T ) = uo(ti) + I r b(zi(s) , s)ds, 0

where by z = Zi ( t ) we denote the characteristic curve through ( z i , ~ ) . From the above inequalities, we get

539

Abbreviations of Universities in This Book

Cincinnati

Columbia

Courant Inst.

Harvard

Illinois

Indiana

Indiana-Purdue

Iowa

Minnesota

Rutgers

Stanford

SUNY, Stony Brook

Toronto

UC, Irvine

University of Cincinnati

Columbia University

Courant Institute of Mathematical Sciences,

New York University

Harvard University

University of Illinois (Urbana)

Indiana University (Bloomington)

Indiana University-Purdue University (Indianapolis)

University of Iowa (Iowa City)

University of Minnesota (Minneapolis)

Rutgers University

Stanford University

State University of New York (Stony Brook)

University of Toronto

University of California (Irvine)

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7350399 Problems and Solutions in Mathematics Li Ta Sien World Scientific

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